35. It is known that $a , b \in \mathrm { R }$ and the complex number $z = a + 2 b \mathrm { i }$ (where i is an imaginary unit) satisfies $z \cdot \bar { z } = 4$, give the following conclusions: (1) The range of $a ^ { 2 } + b ^ { 2 }$ is $[ 1,4 ]$.
The range of values of $a ^ { 2 } + b ^ { 2 }$ is $[ 1,4 ]$; (2) the range of values of $\sqrt { ( a - \sqrt { 3 } ) ^ { 2 } + b ^ { 2 } } + \sqrt { ( a + \sqrt { 3 } ) ^ { 2 } + b ^ { 2 } } = 4$; (3) the range of values of $\frac { b - \sqrt { 5 } } { a }$
is $( - \infty , - 1 ] \cup [ 1 , + \infty )$; (4) The minimum value of $\frac { 1 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } }$ is 2; the number of correct conclusions is ( )
- A. A. 1
- B. B. 2
- C. C. 3
- D. D. 4
Answer: C
Solution: From ${ } ^ { Z \cdot \bar { z } } = 4 \Rightarrow ( a + 2 b \mathrm { i } ) ( a - 2 b \mathrm { i } ) = a ^ { 2 } + 4 b ^ { 2 } = \left. z \right| ^ { 2 } = 4 \Rightarrow \frac { a ^ { 2 } } { 4 } + b ^ { 2 } = 1$, the point $( a , b ) _ { \text {的轨迹 } }$
is an ellipse with $( - \sqrt { 3 } , 0 ) , ( \sqrt { 3 } , 0 )$ as the focus, $a ^ { \prime } = 2$ as the length of the long semiaxis, $b ^ { \prime } = 1$ as the length of the short semiaxis, and $c ^ { \prime } = \sqrt { 3 }$ as the semifocal distance.
From the definition of ellipse, (2) is correct;
$a ^ { 2 } + b ^ { 2 }$ denotes the square of the distance from the point on the ellipse to the origin, it is easy to know that the distance from the endpoints on the short axis of the ellipse to the origin is the smallest, and the distance from the endpoints on the long axis to the origin is the largest, which are 1 and 2 respectively, so the range of values of $a ^ { 2 } + b ^ { 2 }$ is ${ } ^ { [ 1,4 ] }$, (1) Correct; $\frac { b - \sqrt { 5 } } { a }$ represents the slope of the line connecting the point $( a , b )$ and the point $( 0 , \sqrt { 5 } )$ on the ellipse, set the straight line $l : b = k a + \sqrt { 5 }$ is tangent to the ellipse, and the equations of the straight line and the ellipse and simplify the equation to: $( 0 , \sqrt { 5 } )$. INLINE_FORMULA_13]], $\left( \frac { 1 } { 4 } + k ^ { 2 } \right) a ^ { 2 } + 2 \sqrt { 5 } k a + 4 = 0$
$\Delta = 20 k ^ { 2 } - 4 \left( 1 + 4 k ^ { 2 } \right) = 0 \Rightarrow k = \pm 1$ , according to the positional relationship between the point and the ellipse can be known, $\frac { b - \sqrt { 5 } } { a }$ of the range of values is $( - \infty , - 1 ] \cup [ 1 , + \infty )$ , (3) is correct ;
According to the meaning of the question, $\frac { 1 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = \frac { \frac { a ^ { 2 } } { 4 } + b ^ { 2 } } { a ^ { 2 } } + \frac { \frac { a ^ { 2 } } { 4 } + b ^ { 2 } } { b ^ { 2 } } = \frac { b ^ { 2 } } { a ^ { 2 } } + \frac { a ^ { 2 } } { 4 b ^ { 2 } } + \frac { 5 } { 4 } \geq 2 \sqrt { \frac { b ^ { 2 } } { a ^ { 2 } } \times \frac { a ^ { 2 } } { 4 b ^ { 2 } } } + \frac { 5 } { 4 } = \frac { 9 } { 4 }$, when and only when $\frac { b ^ { 2 } } { a ^ { 2 } } = \frac { a ^ { 2 } } { 4 b ^ { 2 } } \Rightarrow a ^ { 2 } = 2 b ^ { 2 } = \frac { 4 } { 3 }$ take "$=$", (4) is wrong.