Hyperbola 1 Question 15
16.
An ellipse intersects the hyperbola $2 x^{2}-2 y^{2}=1$ orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then
(2009)
(a) equation of ellipse is $x^{2}+2 y^{2}=2$
(b) the foci of ellipse are $( \pm 1,0)$
(c) equation of ellipse is $x^{2}+2 y^{2}=4$
(d) the foci of ellipse are $( \pm \sqrt{2}, 0)$
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Answer:
Correct Answer: 16. (a,b)
Solution:
- Given,
$ 2 x^{2}-2 y^{2}=1 $
$\Rightarrow \quad \frac{x^{2}}{\frac{1}{2}}-\frac{y^{2}}{\frac{1}{2}}=1$
Eccentricity of hyperbola $=\sqrt{2}$
So, eccentricity of ellipse $=1 / \sqrt{2}$
Let equation of ellipse be
$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
[where $a>b$ ]
$ \begin{aligned} & \therefore & \frac{1}{\sqrt{2}} & =\sqrt{1-\frac{b^{2}}{a^{2}}} \\ & \Rightarrow & \frac{b^{2}}{a^{2}} & =\frac{1}{2} \Rightarrow a^{2}=2 b^{2} \\ & \Rightarrow & x^{2}+2 y^{2} & =2 b^{2} \end{aligned} $
Let ellipse and hyperbola intersect at
$ A \frac{1}{\sqrt{2}} \sec \theta, \frac{1}{\sqrt{2}} \tan \theta $
On differentiating Eq. (i), we get
$ \begin{aligned} 4 x-4 y \frac{d y}{d x} & =0 \Rightarrow \frac{d y}{d x}=\frac{x}{y} \\ \left.\frac{d y}{d x}\right| _{\text {at } A} & =\frac{\sec \theta}{\tan \theta}=\operatorname{cosec} \theta \end{aligned} $
and on differentiating Eq. (ii), we get
$ 2 x+4 y \frac{d y}{d x}=\left.0 \Rightarrow \frac{d y}{d x}\right| _{\text {at } A}=-\frac{x}{2 y}=-\frac{1}{2} \operatorname{cosec} \theta $
Since, ellipse and hyperbola are orthogonal.
$\therefore \quad-\frac{1}{2} \operatorname{cosec}^{2} \theta=-1 \Rightarrow \operatorname{cosec}^{2} \theta=2 \Rightarrow \theta= \pm \frac{\pi}{4}$
$\therefore \quad A 1, \frac{1}{\sqrt{2}}$ or $1,-\frac{1}{\sqrt{2}}$
Form Eq. (ii), $\quad 1+2 \frac{1}{\sqrt{2}}^{2}=2 b^{2}$
$\Rightarrow \quad b^{2}=1$
Equation of ellipse is $x^{2}+2 y^{2}=2$.
Coordinates of foci $( \pm a e, 0)= \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0=( \pm 1,0)$
If major axis is along $Y$-axis, then
$ \begin{aligned} \frac{1}{\sqrt{2}} & =\sqrt{1-\frac{a^{2}}{b^{2}}} \Rightarrow b^{2}=2 a^{2} \\ \therefore \quad 2 x^{2}+y^{2} & =2 a^{2} \Rightarrow Y^{\prime}=-\frac{2 x}{y} \\ \Rightarrow \quad y _{\frac{1}{\sqrt{2}} \sec \theta, \frac{1}{\sqrt{2}} \tan \theta} & =\frac{-2}{\sin \theta} \end{aligned} $
As ellipse and hyperbola are orthogonal
$ \begin{aligned} & \therefore \quad-\frac{2}{\sin \theta} \cdot \operatorname{cosec} \theta=-1 \\ & \Rightarrow \quad \operatorname{cosec}^{2} \theta=\frac{1}{2} \Rightarrow \theta= \pm \frac{\pi}{4} \\ & \therefore \quad 2 x^{2}+y^{2}=2 a^{2} \\ & \Rightarrow \quad 2+\frac{1}{2}=2 a^{2} \Rightarrow a^{2}=\frac{5}{4} \\ & \therefore \quad 2 x^{2}+y^{2}=\frac{5}{2} \text {, corresponding foci are }(0, \pm 1) \end{aligned} $
Hence, option (a) and (b) are correct.
