If the foci of the ellipse frac{x^2}{16} + fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,we rewrite it as $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$.Here,$a^2 = \fra

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) For the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,we rewrite it as $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$.Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.The eccentricity $e_h$ is given by $e_h^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{81/25}{144/25} = 1 + \frac{81}{144} = \frac{225}{144}$.So,$e_h = \frac{15}{12} = \frac{5}{4}$.The foci of the hyperbola are $(\pm a_h e_h, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$.For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm ae, 0)$ where $a^2 = 16$.Since the foci are the same,$ae = 3$,so $e = \frac{3}{4}$.For an ellipse,$e^2 = 1 - \frac{b^2}{a^2}$,so $(\frac{3}{4})^2 = 1 - \frac{b^2}{16}$.$\frac{9}{16} = 1 - \frac{b^2}{16} \implies \frac{b^2}{16} = 1 - \frac{9}{16} = \frac{7}{16}$.Therefore,$b^2 = 7$.

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

Similar Questions

The major product of the following reaction is

$A$ particle carrying a charge equal to $100$ times the charge on an electron is rotating per second in a circular path of radius $0.8 \ m$. The value of the magnetic field produced at the centre will be (where ${\mu _0}$ is the permeability of vacuum).

Which one feature is common to leech,cockroach,and scorpion?

Resin and turpentine are obtained from

The major product of the following reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module