Test Index

Hyperbola

Section: Mathematics

© examsnet.com

Question : 43 of 98

Marks: +1, -0

For the hyperbola

H: x^2-y^2=1

and the ellipse

E: \;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1, a > b > 0

(1) eccentricity of

E

be reciprocal of the eccentricity of

H

y=\sqrt{\;\frac{5}{2}} x + K

be a common tangent of

F

H

4(a^2+b^2)

is equal to ________.

[28-Jul-2022-Shift-1]

Your Answer:

Solution:

The equation of tangent to hyperbola

x^2-y^2=1

within slope

m

is equal to

y=m x \pm \sqrt{m^2-1}

.

And for same slope

m

, equation of tangent to ellipse

\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1

is

y=m x \pm \sqrt{a^2 m^2+b^2} \dots

. (ii)

\because

Equation (i) and (ii) are identical

\therefore a^2 m^2+b^2=m^2-1

\therefore m^2=\;\frac{1+b^2}{1-a^2}

But equation of common tangent is

y=\sqrt{\;\frac{5}{2}} x+k

\therefore m=\sqrt{\;\frac{5}{2}} \Rightarrow \;\frac{5}{2}=\;\frac{1+b^2}{1-a^2}

\therefore 5 a^2+2 b^2=3 \dots \;\text{(i)}\;

\;\text{eccentricity of ellipse}\;=\;\frac{1}{\sqrt{2}}

\therefore 1-\;\frac{b^2}{a^2}=\;\frac{1}{2}

\Rightarrow a^2=2 b^2 \dots \;\text{(ii)}\;

From equation (i) and (ii) :

a^2=\;\frac{1}{2}, b^2=\;\frac{1}{4}

\therefore 4(a^2+b^2)=3

© examsnet.com

Go to Question:

Prev Question

Next Question