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

Answer: (B)

7

Answer 2

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 $e^2=1-\frac{b^2}{a^2}$
⇒$\frac{9}{16}=1-\frac{b^2}{16}$
⇒$b^2=7$