Question

If the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1,b>0$ and the hyperbola $\frac{x^{2}}{81}-\frac{y^{2}}{63}=\frac{1}{16}$ intersect orthogonally, then the value of $b^{2}$ is

• A. $5$
• B. $7$
• C. $9$
• D. $\frac{81}{7}$

Answer 1

Answer: (B)

If the ellipse and hyperbola are intersecting orthogonally then foci of the two curves are coincident. Now,
$\frac{x^{2}}{81}-\frac{y^{2}}{63}=\frac{1}{16}$
$\Rightarrow \frac{x^{2}}{\frac{81}{16}}-\frac{y^{2}}{\frac{63}{16}}=1$
Comparing with the standard form $\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1$ , we get
$A^{2}=\frac{81}{16},B^{2}=\frac{63}{16}$
Eccentricity of hyperbola is
$E^{2}=1+\frac{B^{2}}{A^{2}}=1+\frac{63}{81}=\frac{144}{81}=\left(\frac{12}{9}\right)^{2}$
$\Rightarrow E^{2}=\left(\frac{12}{9}\right)^{2}\Rightarrow E=\left(\frac{12}{9}\right)=\frac{4}{3}$
Hence, the foci are
$\left(\pm A E , 0\right)\equiv \left(\pm 3 , 0\right)$
Then, eccentricity of an ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1,b>0$ is
$b^{2}=a^{2}\left(1 - e^{2}\right)$
$\Rightarrow b^{2}=a^{2}-a^{2}e^{2}$
$\Rightarrow b^{2}=16-9=7$