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
$\left(\pm AE,0\right)\equiv \left(\pm 3,0\right)$
Then, eccentricity of an ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1,b>0$ is
$b^2=a^2\left(1-e^2\right)$
$\Rightarrow b^2=a^2-a^2{e}^2$
$\Rightarrow b^2=16-9=7$