An ellipse intersects the hyperbola 2 x 2 - 2 y 2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

Question

An ellipse intersects the hyperbola $2 x^{2} - 2 y^{2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

Smart Achievers · Accepted Answer

(1, 2)

The given hyperbola is

$x^{2} - y^{2} = \frac{1}{2} ...(i)$

which is a rectangular hyperbola $(∵ a = b)$ $∴$ $e = \sqrt{2}$

Let the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Its eccentricity $= \frac{1}{\sqrt{2}}$

$∴$ $b^{2} = a^{2} (1 - \frac{1}{2})$ $\Rightarrow b^{2} = \frac{a^{2}}{2}$

Hence, the equation of ellipse becomes

$x^{2} + 2 y^{2} = a^{2} ...(ii)$

Let the hyperbola (i) and ellipse (ii) intersect each other at $P (x_{1}, y_{1})$

Then slope of hyperbola (i) at $P$ is given by

$m_{1} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{x_{1}}{y_{1}}$

and that of ellipse (ii) at $P$ is

$m_{2} = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = - \frac{x_{1}}{2 y_{1}}$

As the two curves intersect orthogonally,

$∴$ $m_{1} m_{2} = - 1$

$\Rightarrow \frac{x_{1}}{y_{1}} (- \frac{x_{1}}{2 y_{1}}) = - 1$ $\Rightarrow x_{1}^{2} = 2 y_{1}^{2} ...(iii)$

Also $P (x_{1}, y_{1})$ lies on $x^{2} - y^{2} = \frac{1}{2}$

$∴$ $x_{1}^{2} - y_{1}^{2} = \frac{1}{2} ...(iv)$

On solving (iii) and (iv), we get $y_{1}^{2} = \frac{1}{2}$ and $x_{1}^{2} = 1$

Also $P (x_{1}, y_{1})$ lies on ellipse $x^{2} + 2 y^{2} = a^{2}$

$∴$ $x_{1}^{2} + 2 y_{1}^{2} = a^{2}$ $\Rightarrow 1 + 1 = a^{2}$ or $a^{2} = 2$

$∴$ Equation of required ellipse is $x^{2} + 2 y^{2} = 2,$ whose foci are $(\pm a e, 0) = (\pm \sqrt{2} \times \frac{1}{\sqrt{2}}, 0) = (\pm 1, 0)$

Solution

Q.
An ellipse intersects the hyperbola $2 x^{2} - 2 y^{2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

1 equation of ellipse is $x^{2} + 2 y^{2} = 2$

2 the foci of ellipse are $(\pm 1, 0)$

3 equation of ellipse is $x^{2} + 2 y^{2} = 4$

4 the foci of ellipse are $(\pm \sqrt{2}, 0)$

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Solution

Q. An ellipse intersects the hyperbola 2x2-2y2=1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then [2009]

1 equation of ellipse is x2+2y2=2

2 the foci of ellipse are (±1,0)

3 equation of ellipse is x2+2y2=4