The foci of the hyperbola 9x^2 - 16y^2 = 144 | vedclass

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

Question

(C) The given equation of the hyperbola is $9x^2 - 16y^2 = 144$. Dividing both sides by $144$,we get $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Comparing t

Vedclass · Accepted Answer

$(\pm 5, 0)$

Vedclass · Answer

(C) The given equation of the hyperbola is $9x^2 - 16y^2 = 144$. Dividing both sides by $144$,we get $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$. The coordinates of the foci are $(\pm ae, 0)$. Substituting the values,we get $(\pm 4 	imes \frac{5}{4}, 0) = (\pm 5, 0)$.

The foci of the hyperbola $9x^2 - 16y^2 = 144$ are

Similar Questions

For what value of $m$ is the line $y = mx + 6$ a tangent to the hyperbola $\frac{x^2}{100} - \frac{y^2}{49} = 1$?

The slope of the tangent drawn from the point $(1,1)$ to the hyperbola $2x^2-y^2=4$ is

The graph of the conic $x^2 - (y - 1)^2 = 1$ has one tangent line with positive slope that passes through the origin. The point of tangency is $(a, b)$. Then the eccentricity of the conic is:

If the tangent at the point $(2 \sec \phi, 3 \tan \phi)$ of the hyperbola $\frac{x^2}{4} - \frac{y^2}{9} = 1$ is parallel to $3x - y + 4 = 0$,then the value of $\phi$ is ............ $^o$.

The eccentricity of the hyperbola $4x^2 - 9y^2 = 16$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module