Let the eccentricity of the hyperbola x2-a2-y2-b2-1 be reciprocal to that of the ellipse x2-9 y2-9- then the ratio a2- b2 equalsA- 8- 1B- 1- 8C- 9- 1D- 1- 9

The correct option is A 8 : 1Equation of the ellipse x^29+y^2=1 is, e = √(1 1/9) = √(8/9) Eccentricity of hyperbola = √(98) So, 1 + b^2a^2 = 98 ⇒b^2a^2 = 18 ...

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Standard XII

Mathematics

Relation between Coefficient and Indices of x and y

Let the eccen...

Question

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse $x^{2} + 9 y^{2} = 9,$ then the ratio $a^{2} : b^{2}$ equals

8 : 1

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1 : 8

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9 : 1

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1 : 9

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Solution

The correct option is A $8 : 1$
Equation of the ellipse
$\frac{x^{2}}{9} + y^{2} = 1$ is,
$e = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}}$
Eccentricity of hyperbola
$= \sqrt{\frac{9}{8}}$
So,
$1 + \frac{b^{2}}{a^{2}} = \frac{9}{8} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{1}{8} \Rightarrow a^{2} : b^{2} = 8 : 1$

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Let the eccentricity of the hyperbola x2a2−y2b2=1 be reciprocal to that of the ellipse x2+9y2=9, then the ratio a2:b2 equals

The correct option is A 8:1Equation of the ellipse x29+y2=1 is, e=√1−19=√89 Eccentricity of hyperbola =√98 So, 1+b2a2=98⇒b2a2=18⇒a2:b2=8:1

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse $x^{2} + 9 y^{2} = 9,$ then the ratio $a^{2} : b^{2}$ equals