Question

Let the ellipse $C_1:\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1(a_1>b_1)$ and the hyperbola $C_2:\frac{x^2}{a_2^2}-\frac{y^2}{b_2^2}=1$ have the same focus point $F_1$ and $F_2$. If point $P$ is the intersection point of $C_1$​ and $C_2$​ in the first quadrant and $|F_1{F}_2|=2|P{F}_2|$, then which of the following is (are) CORRECT?
( $e_1$ and $e_2$ are eccentricities of ellipse and hyperbola respectively.)

• A. $e_1\in (\frac{1}{2},1)$
• B. $e_2-e_1\in (\frac{1}{2},\infty )$
• C. $e_1\in (\frac{1}{4},\frac{1}{2})$
• D. $e_2+e_1\in (\frac{3}{2},\infty )$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $e_1\in (\frac{1}{2},1)$
B $e_2-e_1\in (\frac{1}{2},\infty )$
D $e_2+e_1\in (\frac{3}{2},\infty )$

By the property of the ellipse,
$|P{F}_1|+|P{F}_2|=2a_1\cdots \left(1\right)$
By the property of hyperbola,
$|P{F}_1|-|P{F}_2|=2a_2\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$|P{F}_2|=a_1-a_2$

Given, $|F_1{F}_2|=2a_1{e}_1=2a_2{e}_2$
and $|F_1{F}_2|=2|P{F}_2|$
$\Rightarrow 2a_1{e}_1=2(a_1-a_2)\Rightarrow a_1{e}_1=a_1-\frac{a_1{e}_1}{e_2}\Rightarrow e_1=1-\frac{e_1}{e_2}(\because a_1\ne 0)\Rightarrow e_2=\frac{e_1}{1-e_1}$

We know that,
$e_2>1\Rightarrow \frac{e_1}{1-e_1}>1\Rightarrow e_1>1-e_1(\because 0<e_1<1)\Rightarrow e_1\in (\frac{1}{2},1)$

Now, $e_2=\frac{e_1}{1-e_1}$
Let $e_2=y$ and $e_1=x$
Then, $y=\frac{x}{1-x},x\in (\frac{1}{2},1)$
$y^{\prime }=\frac{1}{(1-x{)}^2}>0,\forall x\in (\frac{1}{2},1)$
So, $y$ is minimum at $x=\frac{1}{2}$ and maximum at $x=1$
$y_{min}=1$ and $y_{max}\to \infty$
$\therefore e_2\in (1,\infty )$
Hence, $e_2+e_1\in (\frac{3}{2},\infty )$
and $e_2-e_1\in (\frac{1}{2},\infty )$