Analytic Geometry

Question 1

Question 1: 1. The point $A ( 1,0 ) , B ( 3,1 )$ is known, and if the line $A B$ is perpendicular to the line $x...

1. The point $A ( 1,0 ) , B ( 3,1 )$ is known, and if the line $A B$ is perpendicular to the line $x - m y + 1 = 0$, then $m =$

A. A. - 2
B. B. $- \frac { 1 } { 2 }$
C. C. $\frac { 1 } { 2 }$
D. D. 2

Answer

Answer: B

Solution: The slope of the line $A B$ is $\frac { 1 - 0 } { 3 - 1 } = \frac { 1 } { 2 }$ according to the question Since the line $A B$ is perpendicular to the line $x - m y + 1 = 0$, the slope of the line $\frac { 1 - 0 } { 3 - 1 } = \frac { 1 } { 2 }$ is $A B$. and the slope of the line $x - m y + 1 = 0$ is - 2 , . So $\frac { 1 } { m } = - 2$, which solves $m = - \frac { 1 } { 2 }$.

Question 2

Question 2: 2. The parabola $y = \frac { 1 } { 2 } x ^ { 2 }$ is known and its collinear equation is

2. The parabola $y = \frac { 1 } { 2 } x ^ { 2 }$ is known and its collinear equation is

A. A. $x = - \frac { 1 } { 8 }$
B. B. $x = \frac { 1 } { 8 }$
C. C. $y = \frac { 1 } { 2 }$
D. D. $y = - \frac { 1 } { 2 }$

Answer

Answer: D

Question 3

Question 3: 3. The parabola $x ^ { 2 } = - 4 y$ has the equation of the collinear line

3. The parabola $x ^ { 2 } = - 4 y$ has the equation of the collinear line

A. A. $x = \frac { 1 } { 16 }$
B. B. $x = 1$
C. C. $y = 1$
D. D. $y = 2$

Answer

Answer: C

Solution: Since $x ^ { 2 } = - 4 y$, which is a parabola with a downward opening, its collinear equation is ${ } ^ { y = 1 }$.

Question 4

Question 4: 4. If the line $y = k x - 2$ is perpendicular to the line $y = 3 x$, then $k =$

4. If the line $y = k x - 2$ is perpendicular to the line $y = 3 x$, then $k =$

A. A. 3
B. B. $\frac { 1 } { 3 }$
C. C. - 3
D. D. $- \frac { 1 } { 3 }$

Answer

Answer: D

Solution: SOLUTION: From two mutually perpendicular lines whose slopes are $\mathrm { k } _ { 1 } , \mathrm { k } _ { 2 } , \mathrm { C } ^ { \mathrm { k } _ { 1 } } \mathrm { x } _ { 2 } = - 1$ can be obtained : $k : 3 = - 1$ , solve for $k = - \frac { 1 } { 3 }$ , .

Question 5

Question 5: 5. The angle of inclination of the line $y = \frac { \sqrt { 3 } } { 3 } x + 7$ is

5. The angle of inclination of the line $y = \frac { \sqrt { 3 } } { 3 } x + 7$ is

A. A. $\frac { \pi } { 3 }$
B. B. $\frac { \pi } { 4 }$
C. C. $\frac { \pi } { 6 }$
D. D. $\frac { \pi } { 2 }$

Answer

Answer: C

Solution: The slope of the straight line $y = \frac { \sqrt { 3 } } { 3 } x + 7$ is $\frac { \sqrt { 3 } } { 3 }$. Let the angle of inclination be $\alpha ( 0 \leq \alpha < \pi )$, then $\tan \alpha = \frac { \sqrt { 3 } } { 3 }$ $\therefore \alpha = \frac { \pi } { 6 }$.

Question 6

Question 6: 7. If two circles $x ^ { 2 } + y ^ { 2 } + 2 \sqrt { m } x + m - 4 = 0 ( m > 0 )$ and $x ^ { 2 } + y...

7. If two circles $x ^ { 2 } + y ^ { 2 } + 2 \sqrt { m } x + m - 4 = 0 ( m > 0 )$ and $x ^ { 2 } + y ^ { 2 } - 4 \sqrt { n } y - 1 + 4 n = 0 ( n > 0 )$ have exactly three common tangents lines, then the minimum value of $\frac { 1 } { m } + \frac { 1 } { n }$ is

A. A. $\frac { 1 } { 9 }$
B. B. $\frac { 4 } { 9 }$
C. C. 1
D. D. 3

Answer

Answer: C

Solution: SOLUTION: From the question, the two circles are tangent to each other The standard equations of the two circles are $( x + \sqrt { m } ) ^ { 2 } + y ^ { 2 } = 4 , x ^ { 2 } + ( y - 2 \sqrt { n } ) ^ { 2 } = 1$ The center of the circle is $( - \sqrt { m } , 0 ) , ( 0,2 \sqrt { n } )$ and the radii are 2 and 1 respectively. Therefore, we have $\sqrt { m + 4 n } = 3$, $\sqrt { m + 4 n } = 3$ $\therefore m + 4 n = 9$ $\therefore \frac { m + 4 n } { 9 } = 1$ $\therefore \frac { 1 } { m } + \frac { 1 } { n } = \frac { m + 4 n } { 9 m } + \frac { m + 4 n } { 9 n }$ $= \frac { 1 } { 9 } + \frac { 4 } { 9 } + \frac { 4 n } { 9 m } + \frac { m } { 9 n } \geq \frac { 5 } { 9 } + 2 \sqrt { \frac { 4 } { 81 } } = 1$ $\frac { 4 n } { 9 m } = \frac { m } { 9 n }$ if and only if $m = 2 n = 3 ^ { \text {when equality holds.} }$

Question 7

Question 7: 8. A social practice group in the research found a stone single-hole bridge (Figure), the bridge par...

8. A social practice group in the research found a stone single-hole bridge (Figure), the bridge parabolic arch part of the bridge span of 21.6 m, the arch from the water 10.9 m, the road thickness of about 1 m. The bridge is a parabolic arch part of the bridge span of 21.6 m, the top of the arch from the water 10.9 m, about 1 m thickness. If the group plans to use a rope from the stone railing of the bridge to lower the camera to get a view so that it falls at the focus of the parabola, the most appropriate length of the rope is ![](/images/questions/analytic-geometry/image-001.jpg)

A. A. 3 m
B. B. 4 m
C. C. 5 m
D. D. 6 m

Answer

Answer: B

Solution: Establish a plane right-angle coordinate system with the vertex of the arch section as the coordinate origin and the horizontal line as the $x$ axis, perpendicular to the $x$ axis, and in the upward direction. ![](/images/questions/analytic-geometry/image-002.jpg) Let the equation of the parabola be $x ^ { 2 } = - 2 p y ( p > 0 )$. It is easy to know that the parabola passes through the point $( 10.8 , - 10.9 )$, then $10.8 ^ { 2 } = 21.8 p$, and we get $p = \frac { 10.8 ^ { 2 } } { 21.8 }$. so $\frac { p } { 2 } = \frac { 5.4 ^ { 2 } } { 10.9 } \approx 2.7$, so $\frac { p } { 2 } + 1 \approx 3.7$.

Question 8

Question 8: 9. If the equation $\frac { x ^ { 2 } } { m } + \frac { y ^ { 2 } } { 2 - m } = 1$ represents a hype...

9. If the equation $\frac { x ^ { 2 } } { m } + \frac { y ^ { 2 } } { 2 - m } = 1$ represents a hyperbola, then the range of values of the real number $m$ is

A. A. $( 0,2 )$
B. B. $( 0,1 ) \cup ( 1,2 )$
C. C. $( - \infty , 0 ) \cup ( 2 , + \infty )$
D. D. $( 2 , + \infty )$

Answer

Answer: C

Solution: If the equation $\frac { x ^ { 2 } } { m } + \frac { y ^ { 2 } } { 2 - m } = 1$ represents a hyperbola, then $m ( 2 - m ) < 0$ , solve for $m < 0$ or $m > 2$ , the That is, the real number $m$ is in the range $( - \infty , 0 ) \cup ( 2 , + \infty )$.

Question 9

Question 9: 10. Let the length of the real axis and the focal length of the hyperbola $C : \frac { x ^ { 2 } } {...

10. Let the length of the real axis and the focal length of the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ be 2,4 respectively, then the asymptote of the hyperbola $C$ is equation is

A. A. $y = \pm \frac { \sqrt { 3 } } { 3 } x$
B. B. $y = \pm \frac { 1 } { 3 } x$
C. C. $y = \pm \sqrt { 3 } x$
D. D. $y = \pm 3 x$

Answer

Answer: C

Solution: Since $2 a = 2,2 c = 4$, the asymptotic equation of $a = 1 , c = 2 , b = \sqrt { 3 }$ is $C$.

Question 10

Question 10: 11. It is known that the hyperbola $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$ , th...

11. It is known that the hyperbola $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$ , then the coordinates of the foci of the hyperbola are

A. A. $( - \sqrt { 7 } , 0 ) , ( \sqrt { 7 } , 0 )$
B. B. $( - 5,0 ) , ( 5,0 )$
C. C. $( 0 , - 5 ) , ( 0,5 )$
D. D. $( 0 , - \sqrt { 7 } ) , ( 0 , \sqrt { 7 } )$

Answer

Answer: B

Solution: According to the hyperbolic equation $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$, the focus is on the $x$ axis, $a ^ { 2 } = 16 , b ^ { 2 } = 9$ $\therefore c ^ { 2 } = a ^ { 2 } + b ^ { 2 } = 16 + 9 = 25$, i.e. ${ } ^ { c = 5 }$. Therefore, the focus is $( - 5,0 ) , ( 5,0 )$.

Question 11

Question 11: 12. Parallel lines \$$l _ { 1 } : 2 x - y + 2 = 0 \text { and }$ l _ { 2 } : m x - y - 3 = 0$ is th...

12. Parallel lines \$$l _ { 1 } : 2 x - y + 2 = 0 \text { and }$ l _ { 2 } : m x - y - 3 = 0$ is the distance between the

A. A. 5
B. B. $\frac { 1 } { 5 }\$
C. C. $\sqrt { 5 }$
D. D. $\frac { \sqrt { 5 } } { 5 }$

Answer

Answer: C

Solution: Since $l _ { 1 } \| l _ { 2 }$, ${ } ^ { 2 \times ( - 1 ) - ( - 1 ) \times m = 0 }$, solve for $m = 2$ and the distance between $l _ { 1 } , l _ { 2 }$ is $\frac { | 2 + 3 | } { \sqrt { 2 ^ { 2 } + ( - 1 ) ^ { 2 } } } = \sqrt { 5 }$.

Question 12

Question 12: 13. For the circle ${ } ^ { C : } x ^ { 2 } + y ^ { 2 } - 4 x + 1 = 0$, the following statements are...

13. For the circle ${ } ^ { C : } x ^ { 2 } + y ^ { 2 } - 4 x + 1 = 0$, the following statements are correct ( )

A. A. Inside the circle ${ } ^ { A ( 1 , - 1 ) }$ of $C$.
B. B. The center of circle $C$ is $( - 2,0 )$.
C. C. The circle $C$ has a radius of 3.
D. D. The circle $C$ is tangent to the line $y = 3$.

Answer

Answer: A

Solution: For A, substituting the point $A ( 1 , - 1 )$ into the circle $C$ yields $1 ^ { 2 } + ( - 1 ) ^ { 2 } - 4 \times 1 + 1 = - 1 < 0$, so the point $A ( 1 , - 1 )$ is in the interior of the circle $C$, and so A is correct; For $\mathrm { B } , \mathrm { C }$, from $x ^ { 2 } + y ^ { 2 } - 4 x + 1 = 0$, we get $( x - 2 ) ^ { 2 } + y ^ { 2 } = 3$, so the center of circle $C$ is $( 2,0 )$, and the radius is $r = \sqrt { 3 }$. FORMULA_10]], so B and C are wrong; For D, the distance from the center of circle $C ( 2,0 ) _ { \text {to the line } } y = 3$ is $d = | 3 - 0 | = 3$, so $3 > \sqrt { 3 }$, that is, $d > r$, and the circle $C _ { \text {and直线 } } { } ^ { y = 3 }$ is away from each other, so D is wrong. D. Error.

Question 13

Question 13: 14. A line passing through the focus $y ^ { 2 } = 4 x$ of the parabola $F$ intersects the parabola a...

14. A line passing through the focus $y ^ { 2 } = 4 x$ of the parabola $F$ intersects the parabola at two points $A , B$, and $M$ is the mid-point of the line segment $A B$. INLINE_FORMULA_5]] is the diameter of the circle must be ( )

A. A. pass through the origin
B. B. Passing point $( - 1,0 )$
C. C. Tangent to the line $x = - 1$.
D. D. Tangent to the line $y = - 1$.

Answer

Answer: C

Solution: Let $A \left( x _ { 1 } , y _ { 1 } \right) , B \left( x _ { 2 } , y _ { 2 } \right)$ , using the focal radius formula we can get: $| A B | = x _ { 1 } + x _ { 2 } + p$ , and $^ { M \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } , \frac { y _ { 1 } + y _ { 2 } } { 2 } \right) }$ , so the distance from $M$ to the line $x = - 1$ is $d = \frac { x _ { 1 } + x _ { 2 } + p } { 2 } = \frac { 1 } { 2 } | A B |$ , so the circle with the line segment $A B$ as the diameter must be a straight line $x = - 1$. INLINE_FORMULA_5]], so the circle with diameter $A B$ is tangent to the line $x = - 1$.

Question 14

Question 14: 15. The following statements are false ( )

15. The following statements are false ( )

A. A. The intercept of the straight line $y = 5 x - 3$ on the $^ { y }$ axis is ${ } ^ { - 3 }$.
B. B. A direction vector of the line $\sqrt { 3 } x - y + 1 = 0$ is $( - \sqrt { 3 } , - 3 )$
C. C. Passes through the point ${ } ^ { ( 3,4 ) }$ and at $x , y _ { \text {轴上的截距相等的直线方程为 } } x + y - 7 = 0$
D. D. $A ( 1,3 ) , B ( 2,5 ) , C ( - 2 , - 3 )$ Three-point common line

Answer

Answer: C

Question 15

Question 15: 16. The equation $x ^ { 2 } + y ^ { 2 } + m x - 2 y + m + 4 = 0$ represents a circle, then the range...

16. The equation $x ^ { 2 } + y ^ { 2 } + m x - 2 y + m + 4 = 0$ represents a circle, then the range of values of $m$ is ( )

A. A. $( - \infty , - 2 ) \cup ( 6 , + \infty )$
B. B. $( - 2,6 )$
C. C. $( - ¥ , - 6 ) \cup ( 2 , + ¥ )$
D. D. $( - 6,2 )$

Answer

Answer: A

Solution: From the question, we can get $m ^ { 2 } + ( - 2 ) ^ { 2 } - 4 ( m + 4 ) > 0$, which is $m ^ { 2 } - 4 m - 12 > 0$ and solve $m < - 2$ or $m > 6$. So choose: A

Question 16

Question 16: 17. The standard equation of the ellipse $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1...

17. The standard equation of the ellipse $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$ that has the same focus as the ellipse $( 5,3 )$ and passes through the point $( 5,3 )$ is ( ).

A. A. $\frac { x ^ { 2 } } { 24 } + \frac { y ^ { 2 } } { 40 } = 1$
B. B. $\frac { x ^ { 2 } } { 40 } + \frac { y ^ { 2 } } { 24 } = 1$
C. C. $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 20 } = 1$
D. D. $\frac { x ^ { 2 } } { 40 } + \frac { y ^ { 2 } } { 26 } = 1$

Answer

Answer: B

Solution: SOLUTION: According to the question, the ellipse $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$ has foci at $( - 4,0 )$ and $( 4,0 )$, which requires that the ellipse's foci are at $x$ axis and ${ } ^ { c = 4 }$; and from the fact that the ellipse passes through the points $( 5,3 )$ then we have $2 a = \sqrt { 81 + 9 } = \sqrt { 1 + 9 } = 4 \sqrt { 10 }$ and $a = 2 \sqrt { 10 }$. then $b = \sqrt { a ^ { 2 } - c ^ { 2 } } = \sqrt { 40 - 16 } = \sqrt { 24 }$; Therefore, the standard equation of the required ellipse is $\frac { x ^ { 2 } } { 40 } + \frac { y ^ { 2 } } { 24 } = 1$;

Question 17

Question 17: 18. If the distance from a point on the hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ {...

18. If the distance from a point on the hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ to the focus $( - \sqrt { 5 } , 0 )$ is greater than the distance from the focus $( \sqrt { 5 } , 0 )$ $b$, then the equation of the hyperbola is ( )

A. A. $\frac { x ^ { 2 } } { 4 } - y ^ { 2 } = 1$
B. B. $\frac { x ^ { 2 } } { 2 } - y ^ { 2 } = 1$
C. C. $x ^ { 2 } - \frac { y ^ { 2 } } { 2 } = 1$
D. D. $x ^ { 2 } - \frac { y ^ { 2 } } { 4 } = 1$

Answer

Answer: D

Solution: From the question, we know that $c = \sqrt { 5 }$ , according to the question, by the definition of hyperbola, we know that $2 a = b$ , and $a ^ { 2 } + b ^ { 2 } = c ^ { 2 }$ , the So $5 a ^ { 2 } = 5$ , we get $a ^ { 2 } = 1 , b ^ { 2 } = 4$ , so the equation of the hyperbola is $x ^ { 2 } - \frac { y ^ { 2 } } { 4 } = 1$ .

Question 18

Question 18: 19. "$0 < a < 3$" is the "[INLINE_FORMULA_1]] hyperbola $\frac { x ^ { 2 } } { a } - \frac { y ^ { 2...

19. "$0 < a < 3$" is the "[INLINE_FORMULA_1]] hyperbola $\frac { x ^ { 2 } } { a } - \frac { y ^ { 2 } } { 9 } = 1 ( a > 0 )$ with centroid greater than 2" ( )

A. A. sufficient and unnecessary condition (math.)
B. B. necessary but insufficient condition (math.)
C. C. necessary and sufficient condition
D. D. Neither sufficient nor necessary condition

Answer

Answer: C

Solution: If the eccentricity of the hyperbola $\frac { x ^ { 2 } } { a } - \frac { y ^ { 2 } } { 9 } = 1 ( a > 0 )$ is greater than ${ } _ { 2 }$, then $\left\{ \begin{array} { l } e = \sqrt { \frac { a + 9 } { a } } > 2 \\ a > 0 \end{array} \right.$, solving for ${ } _ { 0 < a < 3 }$. So " $0 < a < 3$" is a sufficient condition for "the centroid of hyperbola $\frac { x ^ { 2 } } { a } - \frac { y ^ { 2 } } { 9 } = 1 ( a > 0 )$ is greater than 2";

Question 19

Question 19: 20. The distance from the center of the circle $C : x ^ { 2 } + y ^ { 2 } - 2 y = 1$ to the asymptot...

20. The distance from the center of the circle $C : x ^ { 2 } + y ^ { 2 } - 2 y = 1$ to the asymptote of the hyperbola $E : \frac { x ^ { 2 } } { 4 } - y ^ { 2 } = 1$ is ( ).

A. A. $\frac { \sqrt { 5 } } { 5 }$
B. B. $\frac { 2 \sqrt { 5 } } { 5 }$
C. C. $\frac { 3 \sqrt { 5 } } { 5 }$
D. D. $\frac { 4 \sqrt { 5 } } { 5 }$

Answer

Answer: B

Solution: The center $( 0,1 )$ of the circle $C : x ^ { 2 } + y ^ { 2 } - 2 y = 1$. the asymptote $y = \pm \frac { 1 } { 2 } x$ of the hyperbola $E : \frac { x ^ { 2 } } { 4 } - y ^ { 2 } = 1$, i.e. $x \pm 2 y = 0$ So the distance from the center of the circle to the asymptote is $\frac { | 2 | } { \sqrt { 5 } } = \frac { 2 \sqrt { 5 } } { 5 }$.

Question 20

Question 20: 21. The angle of inclination of the line ${ } ^ { x + \sqrt { 3 } y + 2 = 0 }$ is ( ).

21. The angle of inclination of the line ${ } ^ { x + \sqrt { 3 } y + 2 = 0 }$ is ( ).

A. A. $150 ^ { \circ }$
B. B. $120 ^ { \circ }$
C. C. $60 ^ { \circ }$
D. D. $- 30 ^ { \circ }$

Answer

Answer: A

Solution: The slope of the line $x + \sqrt { 3 } y + 2 = 0$ is $k = - \frac { 1 } { \sqrt { 3 } } = - \frac { \sqrt { 3 } } { 3 }$ from the question So the angle of inclination of the line $x + \sqrt { 3 } y + 2 = 0$ is $150 ^ { \circ }$.

Question 21

Question 21: 22. If the chord of a line ${ } ^ { x - y - 2 = 0 }$ intercepted by a circle $^ { ( x - a ) ^ { 2 } ...

22. If the chord of a line ${ } ^ { x - y - 2 = 0 }$ intercepted by a circle $^ { ( x - a ) ^ { 2 } + y ^ { 2 } = 4 }$ is ${ } ^ { 2 \sqrt { 2 } }$, then the real number $a _ { \text {的值为( )} }$ is $a _ { \text {的值为( )} }$.

A. A. - 1 or 3
B. B. 1 or 3
C. C. 0 or 4
D. D. - 2 or 6

Answer

Answer: C

Solution: SOLUTION: From the circle $( x - a ) ^ { 2 } + y ^ { 2 } = 4$, get the center $( a , 0 )$, radius $r = 2$, then the distance from the center of the circle to the straight line $d = \frac { | a - 2 | } { \sqrt { 2 } }$, and the length of the chord $= 2 \sqrt { r ^ { 2 } - d ^ { 2 } } = 2 \sqrt { 4 - \frac { ( a - 2 ) ^ { 2 } } { 2 } } = 2 \sqrt { 2 }$, and solve for [$( a , 0 )$.

Question 22

Question 22: It is known that the left and right foci of the hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac...

It is known that the left and right foci of the hyperbola $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ are $F _ { 1 } , F _ { 2 } , P$ on the right branch of the hyperbola. $P F _ { 2 } \perp F _ { 1 } F _ { 2 } , O H \perp P F _ { 1 } 于 _ { H } , O H = \frac { 1 } { 3 } O F _ { 1 }$ is a point on the right branch of the hyperbola, then the asymptotic equation of the hyperbola is

A. A. $y = \pm 4 x$
B. B. $y = \pm 3 x$
C. C. $y = \pm 2 x$
D. D. $y = \pm x$

Answer

Answer: D

Solution: SOLUTION: When $x = c$, substituting the hyperbola gives $y = \pm \frac { b ^ { 2 } } { a }$, so $\left| P F _ { 2 } \right| = \frac { b ^ { 2 } } { a }$, and $\left| P F _ { 1 } \right| - \left| P F _ { 2 } \right| = 2 a$. so $\left| P F _ { 1 } \right| = 2 a + \frac { b ^ { 2 } } { a }$ Since $P F _ { 2 } \perp F _ { 1 } F _ { 2 } , ~ O H \perp P F _ { 1 }$ , so $\triangle F _ { 1 } O H \sim \triangle F _ { 1 } P F _ { 2 }$ From similar triangles, $\frac { | O H | } { \left| P F _ { 2 } \right| } = \frac { \left| O F _ { 1 } \right| } { \left| P F _ { 1 } \right| }$ because $O H = \frac { 1 } { 3 } O F _ { 1 }$ Therefore $\frac { 1 } { 3 } = \frac { \frac { b ^ { 2 } } { a ^ { 2 } } } { 2 a + \frac { b ^ { 2 } } { a } }$ $\therefore \frac { 2 } { 3 } a ^ { 2 } + \frac { 1 } { 3 } b ^ { 2 } = b ^ { 2 }$, then $a = b$, so the asymptotic equation of the hyperbola is $y = \pm x$;

Question 23

Question 23: 24. If $\odot C _ { 1 } : x ^ { 2 } + y ^ { 2 } - 2 x - 2 y - 14 = 0 , \odot C _ { 2 } : x ^ { 2 } +...

24. If $\odot C _ { 1 } : x ^ { 2 } + y ^ { 2 } - 2 x - 2 y - 14 = 0 , \odot C _ { 2 } : x ^ { 2 } + y ^ { 2 } + 4 x + 6 y + 4 = 0$, then $\oplus C _ { 1 }$ is tangent to $\odot C _ { 2 }$. of $\odot C _ { 2 }$ is

A. A. 1
B. B. 2
C. C. 3
D. D. 4

Answer

Answer: B

Solution: $\odot C _ { 1 } : x ^ { 2 } + y ^ { 2 } - 2 x - 2 y - 14 = 0$. i.e. $\odot C _ { 1 } : ( x - 1 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 16$, center of circle $C _ { 1 } ( 1,1 ) , r _ { 1 } = 4$, $C _ { 1 } ( 1,1 ) , r _ { 1 } = 4$, $\odot C _ { 2 } : x ^ { 2 } + y ^ { 2 } + 4 x + 6 y + 4 = 0$ $\odot C _ { 2 } : x ^ { 2 } + y ^ { 2 } + 4 x + 6 y + 4 = 0$ i.e. $\odot C _ { 2 } : ( x + 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 9$, center of circle $C _ { 2 } ( - 2 , - 3 ) , r _ { 1 } = 3$, $C _ { 2 } ( - 2 , - 3 ) , r _ { 1 } = 3$ then $\left| C _ { 1 } C _ { 2 } \right| = \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } = 5$ , the $\left| C _ { 1 } C _ { 2 } \right| = \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } = 5$ so $r _ { 1 } - r _ { 2 } < 5 < r _ { 1 } + r _ { 2 }$ , and So the two circles intersect and have 2 common tangents.

Question 24

Question 24: 25. The general equation of a circle is known to be $x ^ { 2 } + y ^ { 2 } + 4 x - 2 y - 4 = 0$, the...

25. The general equation of a circle is known to be $x ^ { 2 } + y ^ { 2 } + 4 x - 2 y - 4 = 0$, then the radius of the circle is

A. A. 1
B. B. 2
C. C. 3
D. D. 4

Answer

Answer: C

Solution: The standard equation of a circle is given by $x ^ { 2 } + y ^ { 2 } + 4 x - 2 y - 4 = 0$: $( x + 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 9$, so the radius of the circle is 3.

Question 25

Question 25: 26. It is known that the left and right foci of the ellipse $C : \frac { x ^ { 2 } } { a ^ { 2 } } +...

26. It is known that the left and right foci of the ellipse $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 \quad ( a > b > 0 )$ are $F _ { 1 } , F _ { 2 } , P$ a point on the ellipse. $\angle F _ { 1 } P F _ { 2 } = 60 ^ { \circ }$, if the distance from the origin $O$ to $P F _ { 1 }$ is $\frac { \sqrt { 3 } } { 6 } a$, then the eccentricity of the ellipse is

A. A. $\frac { \sqrt { 2 } } { 5 }$
B. B. $\frac { \sqrt { 6 } } { 3 }$
C. C. $\frac { \sqrt { 7 } } { 3 }$
D. D. $\frac { \sqrt { 3 } } { 3 }$

Answer

Answer: D

Solution: Set $\left| P F _ { 1 } \right| = m , \left| P F _ { 2 } \right| = n$. as ${ } ^ { O N \perp P F _ { 1 } } , \quad F _ { 2 } M \perp P F _ { 1 }$. ![](/images/questions/analytic-geometry/image-003.jpg) From the question, we have $| O N | = \frac { \sqrt { 3 } } { 6 } a , \left| F _ { 2 } M \right| = \frac { \sqrt { 3 } } { 3 } a , \angle F _ { 1 } P F _ { 2 } = 60 ^ { \circ }$ , ${ } ^ { O N \perp P F _ { 1 } } , \quad F _ { 2 } M \perp P F _ { 1 }$ , ![](/images/questions/analytic-geometry/image-003.jpg) That is, we have $| P M | = \frac { 1 } { 3 } a , \left| P F _ { 2 } \right| = \frac { 2 } { 3 } a$, and from $m + n = 2 a$ which leads to $\left| M F _ { 1 } \right| = a$. Since $\left| F _ { 1 } F _ { 2 } \right| = 2 c$, in the right triangle $F _ { 1 } M F _ { 2 }$, by the Pythagorean Theorem, we have $a ^ { 2 } + \left( \frac { \sqrt { 3 } } { 3 } a \right) ^ { 2 } = 4 c ^ { 2 }$ , the which gives $e = \frac { c } { a } = \frac { \sqrt { 3 } } { 3 }$.

Question 26

Question 26: 27. It is known that $a _ { 1 } , a _ { 2 } , b _ { 1 } , b _ { 2 } , c _ { 1 } , c _ { 2 } \in \mat...

27. It is known that $a _ { 1 } , a _ { 2 } , b _ { 1 } , b _ { 2 } , c _ { 1 } , c _ { 2 } \in \mathbf { R }$, the line ${ } ^ { l } : a _ { 1 } x + b _ { 1 } y + c _ { 1 } = 0 , l _ { 2 } : a _ { 2 } x + b _ { 2 } y + c _ { 2 } = 0$ Then "$\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } }$" is the "line $l _ { 1 }$ parallel to $l _ { 2 }$" of $l _ { 2 }$.

A. A. sufficient and unnecessary condition
B. B. necessary but not sufficient condition (math.)
C. C. necessary and sufficient condition
D. D. Neither sufficient nor necessary

Answer

Answer: D

Solution: When $\frac { a _ { 1 } } { b _ { 1 } } = \frac { a _ { 2 } } { b _ { 2 } }$, the two lines may be parallel or coincident, so sufficiency does not hold; When $l _ { 1 } / / l _ { 2 }$, $b _ { 1 }$ and $b _ { 2 }$ may both be equal to 0, so $\frac { a _ { 1 } } { b _ { 1 } } = \frac { a _ { 2 } } { b _ { 2 } }$ does not necessarily hold, and therefore necessity does not hold; In summary, $\frac { a _ { 1 } } { b _ { 1 } } = \frac { a _ { 2 } } { b _ { 2 } }$ is neither sufficient nor necessary for $l _ { 1 } / / l _ { 2 }$.

Question 27

Question 27: 29. It is known that the focus of the parabola $C : y = \frac { 1 } { 4 } x ^ { 2 }$ is $F , P$ is a...

29. It is known that the focus of the parabola $C : y = \frac { 1 } { 4 } x ^ { 2 }$ is $F , P$ is a point on the parabola $C$ and $| P F | = 3$ , then the distance of the point ${ } _ { P }$ to the origin is ( ). $O$ to the origin of the coordinates is ( )

A. A. 2
B. B. $2 \sqrt { 2 }$
C. C. $2 \sqrt { 3 }$
D. D. 4

Answer

Answer: C

Solution: $4 y = x ^ { 2 }$ , then the collinear equation is ${ } ^ { y = - 1 }$ , the Set $P ( m , n )$ , from the question, we can get $n + 1 = 3$ , solve $n = 2$ , then $m ^ { 2 } = 8$ . Therefore, the distance from the point $P$ to the origin $O$ is $\sqrt { m ^ { 2 } + n ^ { 2 } } = \sqrt { 8 + 4 } = 2 \sqrt { 3 }$.

Question 28

Question 28: 30. It is known that the left and right foci of the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } }...

30. It is known that the left and right foci of the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a , b > 0 )$ are $F _ { 1 } ( - c , 0 ) , F _ { 2 } ( c , 0 )$, and if there exists a point $P$ on the left branch of the hyperbola such that $\left| P F _ { 2 } \right| = \frac { 3 } { 2 } C - 2 a$, then the range of values of the eccentricity is ( )

A. A. $( 1,4 ]$
B. B. $[ 6 , + \infty )$
C. C. $[ 4 , + \infty )$
D. D. $[ 2 , + \infty )$

Answer

Answer: B

Question 29

Question 29: 31. Let $F$ be the right focus of the ellipse $\frac { x ^ { 2 } } { 4 } + y ^ { 2 } = 1$, and the m...

31. Let $F$ be the right focus of the ellipse $\frac { x ^ { 2 } } { 4 } + y ^ { 2 } = 1$, and the maximum distance between the point on the ellipse and the point $F$ is $M$, and the minimum distance is $m$, then the coordinates of the point on the ellipse and the point ${ } _ { F }$ are equal to ( ). ], then the coordinates of the point on the ellipse whose distance from the point ${ } _ { F }$ is equal to $\frac { 1 } { 2 } ( M + m )$ are ( )

A. A. $( 0 , \pm 2 )$
B. B. $( 0 , \pm 1 )$
C. C. $\left( \sqrt { 3 } , \pm \frac { 1 } { 2 } \right)$
D. D. $\left( \sqrt { 2 } , \pm \frac { \sqrt { 2 } } { 2 } \right)$

Answer

Answer: B

Solution: SOLUTION: $\because \frac { x ^ { 2 } } { 4 } + y ^ { 2 } = 1$, so $a = 2 , b = 1 , c = \sqrt { a ^ { 2 } - b ^ { 2 } } = \sqrt { 3 }$, the Also the maximum distance between the point on the rough circle and the right focus $F$ is $M$ and the minimum distance is $m$, the $m$ $\therefore M = a + c , n = a - c$. $\therefore \frac { 1 } { 2 } ( M + m ) = a = 2$. Then the points on the ellipse whose distance from the right focus $F$ is equal to $\frac { 1 } { 2 } ( M + m )$ are the two vertices of the short axis with coordinates $( 0 , \pm 1 )$.

Question 30

Question 30: 32. $M \left( x _ { 0 } , y _ { 0 } \right)$ is a point in the circle $x ^ { 2 } + y ^ { 2 } = 1$ th...

32. $M \left( x _ { 0 } , y _ { 0 } \right)$ is a point in the circle $x ^ { 2 } + y ^ { 2 } = 1$ that is different from the center of the circle, then the position of the line $x _ { 0 } x + y _ { 0 } y = 1$ with the circle is

A. A. tangency
B. B. intersect
C. C. leave each other
D. D. Tangent or intersection

Answer

Answer: C

Solution: From the question, $M \left( x _ { 0 } , y _ { 0 } \right)$ is a point inside the circle $x ^ { 2 } + y ^ { 2 } = 1$ which is different from the center of the circle. Then $x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } < 1$, $x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } < 1$ and the distance from the center of the circle $x ^ { 2 } + y ^ { 2 } = 1$ to the line $x _ { 0 } x + y _ { 0 } y = 1$ is $d = \frac { 1 } { \sqrt { x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } } } > 1 = r$. Therefore, the positional relationship between the line $x _ { 0 } x + y _ { 0 } y = 1$ and the circle is disjoint.

Question 31

Question 31: 33. It is known that hyperbola $C : \frac { x ^ { 2 } } { 12 } - \frac { y ^ { 2 } } { 4 } = 1$, poi...

33. It is known that hyperbola $C : \frac { x ^ { 2 } } { 12 } - \frac { y ^ { 2 } } { 4 } = 1$, point $F$ is the right focus of $C$, and point $P$ is the moving point on the left branch of $C$. INLINE_FORMULA_5]] to the left branch of $P$. $C$ to an asymptote of $d$, then the minimum value of $d + | P F |$ is ( )

A. A. $2 + 4 \sqrt { 3 }$
B. B. $6 \sqrt { 3 }$
C. C. 8
D. D. 10

Answer

Answer: A

Solution: From the hyperbola $C : \frac { x ^ { 2 } } { 12 } - \frac { y ^ { 2 } } { 4 } = 1$, it follows that $a = 2 \sqrt { 3 } , b = 2 , F ( 40 )$ Let the left focus of the hyperbola be $F ^ { \prime } ( - 40 )$, and let an asymptote be $l : y = - \frac { b } { a } x = - \frac { \sqrt { 3 } } { 3 } x$, which is $x + \sqrt { 3 } y = 0$. as $P E \perp l$, and the vertical foot is $E$, i.e. $| P E | = d$, and the vertical foot is $| P E | = d$. as $_ { F \prime } H \perp l$ with a plumb bob of $H$, that is $\left| F ^ { \prime } H \right| = \frac { | - 4 | } { \sqrt { 1 ^ { 2 } + ( \sqrt { 3 } ) ^ { 2 } } } = 2$, and ![](/images/questions/analytic-geometry/image-004.jpg) since the point $P$ is a moving point on the left branch of $C$. So $| P F | - \left| P F ^ { \prime } \right| = 2 a$ , which gives $| P F | = 2 a + \left| P F ^ { \prime } \right|$ , the Therefore ${ } ^ { d + | F P | = | P E | + 2 a + \left| P F ^ { \prime } \right| = 2 a + | P E | + \left| P F ^ { \prime } \right| \text { ,} }$ From the figure, $P , F ^ { \prime } , E$ is minimized by $2 \times 2 \sqrt { 3 } + \left| F ^ { \prime } H \right| = 4 \sqrt { 3 } + 2$ when the three points of $P , F ^ { \prime } , E$ are co-linear, i.e., when the points of $E$ and $H$ are coincident. i.e. $d + | P F | _ { \text {has minimum value } } 4 \sqrt { 3 } + 2$.

Question 32

Question 32: 34. Let $F _ { 1 } , F _ { 2 }$ be the left and right foci of the hyperbola $C : x ^ { 2 } - \frac {...

34. Let $F _ { 1 } , F _ { 2 }$ be the left and right foci of the hyperbola $C : x ^ { 2 } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, and the vertical line from $x$ to $C$ over $A$, $B$ and $\triangle A B F _ { 1 }$ is a positive triangle. INLINE_FORMULA_5]], $B$, if $\triangle A B F _ { 1 }$ is a positive triangle, then ( )

A. A. $b = 2$
B. B. $C$ has a focus of $2 \sqrt { 3 }$
C. C. $C$ has a centroid of $\frac { \sqrt { 6 } } { 3 }$
D. D. The area of $\triangle A B F _ { 1 }$ is $2 \sqrt { 2 }$

Answer

Answer: B

Solution: From the hyperbola $C : x ^ { 2 } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, we get $c = \sqrt { 1 + b ^ { 2 } } , ~ \therefore F _ { 1 } \left( - \sqrt { 1 + b ^ { 2 } } , ~ 0 \right) , ~ F _ { 2 } \left( \sqrt { 1 + b ^ { 2 } } , ~ 0 \right)$, and substituting $x = \sqrt { 1 + b ^ { 2 } }$ into the hyperbola equation gives us: $1 + b ^ { 2 } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$, which solves for $y = \pm b ^ { 2 }$. It may be useful to take $A \left( \sqrt { 1 + b ^ { 2 } } , b ^ { 2 } \right) , B \left( \sqrt { 1 + b ^ { 2 } } , - b ^ { 2 } \right)$, $A \left( \sqrt { 1 + b ^ { 2 } } , b ^ { 2 } \right) , B \left( \sqrt { 1 + b ^ { 2 } } , - b ^ { 2 } \right)$ $\triangle A B F _ { 1 }$ to be a square triangle. $\tan \angle A F _ { 1 } F _ { 2 } = \tan 30 ^ { \circ } = \frac { \sqrt { 3 } } { 3 } = \frac { \left| A F _ { 2 } \right| } { \left| F _ { 1 } F _ { 2 } \right| } = \frac { b ^ { 2 } } { 2 \sqrt { 1 + b ^ { 2 } } }$, $A \left( \sqrt { 1 + b ^ { 2 } } , b ^ { 2 } \right) , B \left( \sqrt { 1 + b ^ { 2 } } , - b ^ { 2 } \right)$, $\triangle A B F _ { 1 }$ are square triangles. Solve for $b ^ { 2 } = 2$, $\tan \angle A F _ { 1 } F _ { 2 } = \tan 30 ^ { \circ } = \frac { \sqrt { 3 } } { 3 } = \frac { \left| A F _ { 2 } \right| } { \left| F _ { 1 } F _ { 2 } \right| } = \frac { b ^ { 2 } } { 2 \sqrt { 1 + b ^ { 2 } } }$ $\therefore b = \sqrt { 2 } , ~ c = \sqrt { 3 } , ~ 2 c = 2 \sqrt { 3 } , e = \frac { c } { a } = \sqrt { 3 }$, $\therefore b = \sqrt { 2 } , ~ c = \sqrt { 3 } , ~ 2 c = 2 \sqrt { 3 } , e = \frac { c } { a } = \sqrt { 3 }$, $\therefore b = \sqrt { 2 } , ~ c = \sqrt { 3 } , ~ 2 c = 2 \sqrt { 3 } , e = \frac { c } { a } = \sqrt { 3 }$ $S _ { \triangle A B F _ { 1 } } = \frac { 1 } { 2 } \times 2 c \times 2 b ^ { 2 } = 4 \sqrt { 3 }$.

Question 33

Question 33: 35. It is known that the left and right foci $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } ...

35. It is known that the left and right foci $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ of the hyperbola $F _ { 1 } , F _ { 2 } , c$ are semifocal distances, and $P$ is a point on the hyperbola which is distinct from the endpoints of the real axis and satisfies $c \tan \angle P F _ { 1 } F _ { 2 } = a \tan \angle P F _ { 2 } F _ { 1 }$, then the eccentricity of the hyperbola [[]] is the value of $e$. INLINE_FORMULA_4]] is in the range of

A. A. $( 1 + \sqrt { 2 } , 1 + \sqrt { 3 } )$ B. $( 1 + \sqrt { 2 } , + \infty )$
B. B. $( 1 + \sqrt { 2 } , 1 + \sqrt { 3 } )$ B. $( 1 + \sqrt { 2 } , + \infty )$
C. C. $( \sqrt { 2 } , 1 + \sqrt { 2 } )$
D. D. $( 1,1 + \sqrt { 2 } )$

Answer

Answer: B

Solution: Because $c \tan \angle P F _ { 1 } F _ { 2 } = a \tan \angle P F _ { 2 } F _ { 1 }$. Therefore $e = \frac { c } { a } = \frac { \tan \angle P F _ { 2 } F _ { 1 } } { \tan \angle P F _ { 1 } F _ { 2 } }$ , . Let $P ( m , n ) , F _ { 1 } ( - c , 0 ) , F _ { 2 } ( c , 0 )$ , . so that $e = \frac { \tan \angle P F _ { 2 } F _ { 1 } } { \tan \angle P F _ { 1 } F _ { 2 } } = - \frac { n } { m - c } \cdot \frac { m + c } { n } = - \frac { m + c } { m - c } = - 1 - \frac { 2 c } { m - c }$ , . because $m > a$ , . so $- 1 - \frac { 2 c } { m - c } > - 1 + \frac { - 2 c } { a - c } = - 1 + \frac { 2 e } { e - 1 }$. so $e + 1 > \frac { 2 e } { e - 1 }$ . i.e. $e ^ { 2 } - 2 e - 1 > 0$ . Solve for $e > 1 + \sqrt { 2 }$.

Question 34

Question 34: 36. Let the line $2 x + ( k - 3 ) y - 2 k + 6 = 0$ pass through the fixed point $P$, then the coordi...

36. Let the line $2 x + ( k - 3 ) y - 2 k + 6 = 0$ pass through the fixed point $P$, then the coordinates of the point $P$ are ( ).

A. A. $( 3,0 )$
B. B. $( 0,2 )$
C. C. $( 0,3 )$
D. D. $( 2,0 )$

Answer

Answer: B

Solution: Equate the line to $( 2 x - 3 y + 6 ) + k ( y - 2 ) = 0$. At that time $\left\{ \begin{array} { l } y - 2 = 0 \\ 2 x - 3 y + 6 = 0 \text { 即 } \end{array} \left\{ \begin{array} { l } x = 0 \\ y = 2 \text { ,直线 } 2 x + ( k - 3 ) y - 2 k + 6 = 0 \text { 恒过定点 } ( 0,2 ) \text { ,} \end{array} \right. \right.$

Question 35

Question 35: 37. It is known that the proposition $p :$ that the line $l : y = m x - 2$ passes through the fixed ...

37. It is known that the proposition $p :$ that the line $l : y = m x - 2$ passes through the fixed point $( 0,2 )$, and that the proposition $q : n = 1$ is a necessary condition for the line $l _ { 1 } : x + n y - 1 = 0$ to be perpendicular to the line $l _ { 2 } : y = n x + 1$. FORMULA_5]] is perpendicular to the line $l _ { 2 } : y = n x + 1$, then the following proposition is true

A. A. $p ^ { \wedge } q$
B. B. $p ^ { \wedge } \neg q$
C. C. $\neg p ^ { \wedge } q$
D. D. $\neg p ^ { \vee } q$

Answer

Answer: D

Solution: Solution: Proposition $p$ : The line $^ { l : y = m x - 2 }$ passes through the fixed point $^ { ( 0 , - 2 ) }$, so the proposition $p$ is false, $\neg p$ is true. INLINE_FORMULA_5]] is constant, the line ${ } ^ { l }$ is perpendicular to the line ${ } ^ { l }$ for any ${ } ^ { n \in \mathrm { R } }$, and so the proposition $q : n = 1$ is the line ${ } ^ { n - n } = 0$. INLINE_FORMULA_10]] and the line $l _ { 2 } : y = n x + 1$ are perpendicular to each other, so proposition $q$ is false, $\neg q$ is true, and so [[INLINE_FORMULA_14 ]] are both false propositions and $\neg p ^ { \wedge } q$ is true.

Question 36

Question 36: 38. In the plane rectangular coordinate system $x O y$, $A ( - 3,0 ) , B ( 1,0 ) , P$ is known to be...

38. In the plane rectangular coordinate system $x O y$, $A ( - 3,0 ) , B ( 1,0 ) , P$ is known to be a moving point on the circle $C : ( x - 3 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 1$, then the minimum value of $| P A | ^ { 2 } + | P B | ^ { 2 }$ is ( )

A. A. 34
B. B. 40
C. C. 44
D. D. 48

Answer

Answer: B

Solution: Let $P ( x , y )$, then $| P A | ^ { 2 } + | P B | ^ { 2 } = ( x + 3 ) ^ { 2 } + y ^ { 2 } + ( x - 1 ) ^ { 2 } + y ^ { 2 } = 2 x ^ { 2 } + 2 y ^ { 2 } + 4 x + 10$ $= 2 \left[ ( x + 1 ) ^ { 2 } + y ^ { 2 } \right] + 8$, i.e. $| P A | ^ { 2 } + | P B | ^ { 2 }$ is equivalent to point $| P A | ^ { 2 } + | P B | ^ { 2 }$. i.e. $| P A | ^ { 2 } + | P B | ^ { 2 }$ is equivalent to two times the square of the distance from the point $P$ to the point $Q ( - 1,0 )$ plus eight. and $| P Q | \geq | Q C | - | P C | = \sqrt { ( 3 + 1 ) ^ { 2 } + 3 ^ { 2 } } - 1 = 5 - 1 = 4$. That is, $| P A | ^ { 2 } + | P B | ^ { 2 } \geq 2 \times 4 ^ { 2 } + 8 = 40$.

Question 37

Question 37: 39. It is known that the ellipse $C _ { 1 } : \frac { x ^ { 2 } } { 13 } + y ^ { 2 } = 1$, the hyper...

39. It is known that the ellipse $C _ { 1 } : \frac { x ^ { 2 } } { 13 } + y ^ { 2 } = 1$, the hyperbola $C _ { 2 } : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a , b > 0 )$, if the circle with the long axis of $C _ { 1 }$ as diameter intersects an asymptote of ${ } ^ { C _ { 2 } }$ at two points of $A , B$, and the two intersections of the ellipse ${ } ^ { C _ { 1 } }$ with this asymptote bisect the line segment $A B$ in three equal parts, then $C _ { 2 }$ The eccentricity of the segment $C _ { 2 }$ is

A. A. $\sqrt { 3 }$
B. B. 3
C. C. $\sqrt { 5 }$
D. D. 5

Answer

Answer: A

Solution: Set $O A _ { \text {的方程为 } } y = k x \left( k > 0 , x _ { 0 } > 0 \right)$, $\therefore _ { \text {设 } } A \left( x _ { 0 } , k x _ { 0 } \right)$ $\therefore _ { \text {设 } } A \left( x _ { 0 } , k x _ { 0 } \right)$ . From the known, we have $| O A | = \sqrt { 13 }$ i.e. $\sqrt { 1 + k ^ { 2 } } x _ { 0 } = \sqrt { 13 }$. Solve for $x _ { 0 } = \frac { \sqrt { 13 } } { \sqrt { 1 + k ^ { 2 } } }$. Therefore $A \left( \frac { \sqrt { 13 } } { \sqrt { 1 + k ^ { 2 } } } , \frac { \sqrt { 13 } k } { \sqrt { 1 + k ^ { 2 } } } \right)$, $A \left( \frac { \sqrt { 13 } } { \sqrt { 1 + k ^ { 2 } } } , \frac { \sqrt { 13 } k } { \sqrt { 1 + k ^ { 2 } } } \right)$ $\therefore A B$ has the coordinates of a trinomial of $\left( \frac { \sqrt { 13 } } { 3 \sqrt { 1 + k ^ { 2 } } } , \frac { \sqrt { 13 } k } { 3 \sqrt { 1 + k ^ { 2 } } } \right)$, which is on the ellipse of $\therefore \frac { \left( \frac { \sqrt { 13 } } { 3 \sqrt { 1 + k ^ { 2 } } } \right) ^ { 2 } } { 13 } + \left( \frac { \sqrt { 13 } k } { 3 \sqrt { 1 + k ^ { 2 } } } \right) ^ { 2 } = 1$. i.e. $1 + 13 k ^ { 2 } = 9 \left( 1 + k ^ { 2 } \right)$ . Solving for $k ^ { 2 } = 2$ gives $\frac { b ^ { 2 } } { a ^ { 2 } } = 2 , b ^ { 2 } = 2 a ^ { 2 }$ , which is Solve for $e = \frac { c } { a } = \sqrt { \frac { a ^ { 2 } + b ^ { 2 } } { a ^ { 2 } } } = \sqrt { 3 }$.

Question 38

Question 38: 40. A square $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ has prisms of length 5, a point $M$...

40. A square $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ has prisms of length 5, a point $M$ is on the prisms $A B$ and $A M = 2$, and the point $P$ is a moving point in the lower $A B C D$ of the square, and the moving point $P$ goes to the straight line $A B C D$, including the boundary. INLINE_FORMULA_5]], and the squared difference between the distance of the moving point $P$ to the line ${ } ^ { A _ { 1 } D _ { 1 } }$ and the distance of the point $P$ to the point $M$ is 25 , then the minimum value of the moving point $P$ to the point $B$ is High School Mathematics Assignment, October 29, 2025

A. A. $\frac { 7 } { 2 }$
B. B. $2 \sqrt { 3 }$
C. C. $\sqrt { 6 }$
D. D. $\sqrt { 2 }$

Answer

Answer: B

Solution: ![](/images/questions/analytic-geometry/image-005.jpg) As shown in the figure, make ${ } ^ { P Q \perp A D } , Q$ the vertical foot, then ${ } ^ { P Q \perp }$ the plane $^ { A D D _ { 1 } A _ { 1 } }$, and across the point $Q$ make ${ } ^ { Q R \perp A D }$, intersecting at $R$, then ${ } ^ { A } D _ { 1 } \perp$ is the plane $P Q R$, so $P R$ is $P$ to the line $P$. INLINE_FORMULA_11]]. Since $P R ^ { 2 } - P M ^ { 2 } = 25$ and $P R ^ { 2 } - P Q ^ { 2 } = R Q ^ { 2 } = 25$, $P M = P Q$. So the trajectory of the point $P$ is a parabola with $A D$ as the collinear and the point $M$ as the focus. ![](/images/questions/analytic-geometry/image-006.jpg) If a right-angle coordinate system is established as shown in the figure, the equation of the trajectory of the point $P$ is $y ^ { 2 } = 4 x ( 0 \leq y \leq 4 )$, the point $_ { A ( - 1,0 ) , B ( 4,0 ) }$, and the point $P \left( \frac { y ^ { 2 } } { 4 } , y \right)$ is set to be $P \left( \frac { y ^ { 2 } } { 4 } , y \right)$, so $| P B | = \sqrt { \left( \frac { y ^ { 2 } } { 4 } - 4 \right) ^ { 2 } + y ^ { 2 } } = \sqrt { \frac { y ^ { 4 } } { 16 } - y ^ { 2 } + 16 }$ $= \sqrt { \frac { 1 } { 16 } \left( y ^ { 2 } - 8 \right) ^ { 2 } + 12 }$, so when $y ^ { 2 } = 8 , | P B |$ obtains the minimum value $2 \sqrt { 3 }$.

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