Find the equation of the hyperbola satisfying | vedclass

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

Question

(A) The vertices are $(\pm 7, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given $a = 7$ a

Vedclass · Accepted Answer

$\frac{x^2}{49} - \frac{9y^2}{343} = 1$

Vedclass · Answer

(A) The vertices are $(\pm 7, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given $a = 7$ and eccentricity $e = \frac{c}{a} = \frac{4}{3}$. Substituting $a = 7$,we get $\frac{c}{7} = \frac{4}{3}$,so $c = \frac{28}{3}$. Using the relation $c^2 = a^2 + b^2$,we have $b^2 = c^2 - a^2 = (\frac{28}{3})^2 - 7^2 = \frac{784}{9} - 49 = \frac{784 - 441}{9} = \frac{343}{9}$. Substituting $a^2 = 49$ and $b^2 = \frac{343}{9}$ into the standard equation,we get $\frac{x^2}{49} - \frac{y^2}{343/9} = 1$,which simplifies to $\frac{x^2}{49} - \frac{9y^2}{343} = 1$.

Find the equation of the hyperbola satisfying the given conditions: Vertices $(\pm 7, 0)$,$e = \frac{4}{3}$.

Similar Questions

The locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}t = 0$ and $\sqrt{3}tx + ty - 4\sqrt{3} = 0$ (where $t$ is a parameter) is a hyperbola whose eccentricity is

$A$ rectangular hyperbola of latus rectum $2$ units passes through $(0, 0)$ and has $(1, 0)$ as its one focus. The other focus lies on the curve -

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

If the ratio of the distance between the foci to the distance between the directrices of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $3 : 2$,then $a : b = \dots$

The angle between the tangents drawn from the point $(2\sqrt{2}, 1)$ to the hyperbola $16x^2 - 25y^2 = 400$ is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module