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

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

Question

(B) 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

(B) 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 $b_H^2 = \frac{81}{25}$.The eccentricity $e_H$ is given by $e_H^2 = 1 + \frac{b_H^2}{a^2} = 1 + \frac{81/25}{144/25} = 1 + \frac{81}{144} = \frac{225}{144}$.So,$e_H = \frac{15}{12} = \frac{5}{4}$.The foci are $(\pm ae_H, 0) = (\pm \frac{12}{5} 	imes \frac{5}{4}, 0) = (\pm 3, 0)$.For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm ae_E, 0)$ where $a^2 = 16$.Since the foci coincide,$ae_E = 3$,so $4e_E = 3$,which means $e_E = \frac{3}{4}$.Using $b^2 = a^2(1 - e_E^2)$,we get $b^2 = 16(1 - (\frac{3}{4})^2) = 16(1 - \frac{9}{16}) = 16(\frac{7}{16}) = 7$.

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

Similar Questions

The number of solutions of the system of equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y - 3z = 5$ is

Water rises in a capillary tube of radius $r$ to a height $h$. The mass of water in the capillary is $m$. What will be the mass of water that rises in a capillary tube of radius $\frac{r}{3}$?

If $x > 0$ and $x^y = e^{x-y}$,then $\frac{dy}{dx}$ is equal to

If the slope of one line of $ax^2 + 4xy + y^2 = 0$ is $3$ times the other,then the value of $a$ is:

In a seconds pendulum,the mass of the bob is $30\, g$. If it is replaced by a $90\, g$ mass,then its time period will be ... $\sec$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module