An ellipse (x2/a2)+(y2/b2)=1 and the hyperbola 2 x2-2 y2=-1 intersect orthogonally. If product of eccentricity of the ellipse and hyperbola is 1 , then (a2/b2) is equal to

Question

An ellipse (x2/a2)+(y2/b2)=1 and the hyperbola 2 x2-2 y2=-1 intersect orthogonally. If product of eccentricity of the ellipse and hyperbola is 1 , then (a2/b2) is equal to (A) (1/2) (B) (1/4) (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

((x0/y0))((-b2 x0/a2 y0))=-1 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/baf439f51f4798b76e680dbf03da7a9a-.png" /> ( a 2/ b 2)=( x 02/ y 02)=( x 02/ x 02+(1)2)<1 eccentricity of ellipse e 2=1-( a 2/ b 2)=(1/2) ⇒ ( a 2/ b 2)=(1/2)

Q. An ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ and the hyperbola $2 x^{2} - 2 y^{2} = - 1$ intersect orthogonally. If product of eccentricity of the ellipse and hyperbola is 1 , then $\frac{a ^{2}}{b ^{2}}$ is equal to

$\frac{1}{2}$

$\frac{1}{4}$

2

4

$(\frac{x _{0}}{y _{0}}) (\frac{- b ^{2} x _{0}}{a ^{2} y _{0}}) = - 1$

$\frac{a ^{2}}{b ^{2}} = \frac{x _{0}^{2}}{y _{0}^{2}} = \frac{x _{0}^{2}}{x _{0}^{2} + \frac{1}{2}} < 1$
eccentricity of ellipse
$e^{2} = 1 - \frac{a ^{2}}{b ^{2}} = \frac{1}{2} \Rightarrow \frac{a ^{2}}{b ^{2}} = \frac{1}{2}$

Q. An ellipse a2x2​+b2y2​=1 and the hyperbola 2x2−2y2=−1 intersect orthogonally. If product of eccentricity of the ellipse and hyperbola is 1 , then b2a2​ is equal to

21​

41​

2

4

(y0​x0​​)(a2y0​−b2x0​​)=−1 b2a2​=y02​x02​​=x02​+21​x02​​<1 eccentricity of ellipse e2=1−b2a2​=21​⇒b2a2​=21​

Q. An ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ and the hyperbola $2 x^{2} - 2 y^{2} = - 1$ intersect orthogonally. If product of eccentricity of the ellipse and hyperbola is 1 , then $\frac{a ^{2}}{b ^{2}}$ is equal to

$\frac{1}{2}$

$\frac{1}{4}$