If $e_{1}$ is the eccentricity of the hyperbola xy = 1 and $e_{2}$ is the eccentricity of the ellipse which is confocal with the hyperbola and $e_{1} e_{2} = 1,$ then select the correct option(s)

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Equation of the ellipse is $3 x^{2} + 3 y^{2} - 2 x y - 16 = 0$

Equation of the ellipse is $3 x^{2} + y^{2} + 2 x y - 12 = 0$

Product of abscissa of the point of intersection of ellipse and hyperbola is 1

Product of abscissa of the point of intersection of ellipse and hyperbola is 1/3

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Rectangular hyperbola

Major axis y = x; minor axis x + y = 0

$e_{2} = \frac{1}{\sqrt{2}}$

$a e_{2} = 2 \Rightarrow a = 2 \sqrt{2}$

$b = 2$

$∴ e l l i p s e \frac{{(\frac{x + y}{\sqrt{2}})}^{2}}{8} + \frac{{(\frac{x - y}{\sqrt{2}})}^{2}}{4} = 1$

$∴ e l l i p s e \frac{{(x + y)}^{2}}{16} + \frac{{(x - y)}^{2}}{8} = 1$

Equation of the ellipse is $3 x^{2} + 3 y^{2} - 2 x y - 16 = 0$

$\Rightarrow 3 x^{2} + \frac{3}{x^{2}} - 2 - 16 = 0$

$\Rightarrow 3 x^{4} - 18 x^{2} + 3 = 0$

$\Rightarrow x_{1} x_{2} x_{3} x_{4} = 1$

Watch 3-min video & get full concept clarity