Question

If the foci of the ellipse $ \frac {x^{2}} {{16}}$+$ \frac {y^{2} }{{b}^2} $=1 and hyperbola $\frac {x^{2}} {144} - \frac {y^{2}}{81}=\frac {1}{25}$ coincide then the value of $b^2$ is

• A. 1
• B. 7
• C. 5
• D. 9

Answer 1

The Correct Option is B

The correct answer is B:7
Given that:
Equation of hyperbola is;
\(\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}\)
⇒\(\frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1\)
\(\therefore a^2=\frac{144}{25} , b^2=\frac{81}{25}  \therefore e=\sqrt{1+\frac{b^2}{a^2}}\)
\(\therefore\) Focii is (±ae,0) or (±3,0) \(=\sqrt{1+\frac{81}{144}}=\frac{15}{12}\)
Now for ellipse,
i.e, \(\frac{x^2}{16}+\frac{y^2}{b^2}=1\)
\(a^2=16\) ,then by considering eccentricity ‘e’,focii is (±4e,0)
as focii ellipse and hyperbola coincides each with other
then; 4e=3
⇒\(e=\frac{3}{4}\)
\(\therefore\) for ellipse \(e^2=1-\frac{b^2}{a^2}\)
⇒\(\frac{9}{16}=1-\frac{b^2}{16}\)
⇒\(b^2=7\)