If e1 is the eccentricity of the ellipse (x2/16)+(y2/25)=1 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2=1 then equation of the hyperbola is

Question

If e1 is the eccentricity of the ellipse (x2/16)+(y2/25)=1 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2=1 then equation of the hyperbola is (A)  (x2/9)- (y2/16)=1 (B)  (x2/16)- (y2/9)=-1 (C)  (x2/9)- (y2/25)=1 (D) None of these

Tardigrade Admin · Accepted Answer

The eccentricity of (x2/16)+ (y2/25)=1 is e1=(√1-(16/25))=(3/5) ∴ e2=(5/3) [∵ e1e2=1] ⇒ Foci of ellipse (0,± 3) ⇒ Equation of hyperbola is (x2/16)-(y2/9)=-1.

Q. If $e_{1}$ is the eccentricity of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1$ and $e_{2}$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e_{1} e_{2} = 1$ then equation of the hyperbola is

$\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$

$\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = - 1$

$\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1$

None of these

The eccentricity of $\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1$ is
$e_{1} = (1 - \frac{16}{25}) = \frac{3}{5}$
$∴ e_{2} = \frac{5}{3}$ $[∵ e_{1} e_{2} = 1]$
$\Rightarrow$ Foci of ellipse $(0, \pm 3)$
$\Rightarrow$ Equation of hyperbola is $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = - 1.$

Q. If e1​ is the eccentricity of the ellipse 16x2​+25y2​=1 and e2​ is the eccentricity of the hyperbola passing through the foci of the ellipse and e1​e2​=1 then equation of the hyperbola is

9x2​−16y2​=1

16x2​−9y2​=−1

9x2​−25y2​=1