If the hyperbola frac{x^2}{a^2} - frac{y^2}{b | vedclass

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

Question

(C) The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.Given $e = \frac{5}{4}$,we have $1 + \frac{b^2}{a^2} = \left(\f

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.Given $e = \frac{5}{4}$,we have $1 + \frac{b^2}{a^2} = \left(\frac{5}{4}ight)^2 = \frac{25}{16}$.Thus,$\frac{b^2}{a^2} = \frac{25}{16} - 1 = \frac{9}{16}$,which implies $b^2 = \frac{9}{16} a^2$.The length of the latus rectum is $\frac{2b^2}{a} = 9$.Substituting $b^2 = \frac{9}{16} a^2$ into the latus rectum equation:$\frac{2}{a} \left(\frac{9}{16} a^2ight) = 9$$\frac{9}{8} a = 9 \Rightarrow a = 8$.Now,$b^2 = \frac{9}{16} (8)^2 = \frac{9}{16} 	imes 64 = 36$,so $b = 6$.Therefore,$ab = 8 	imes 6 = 48$.

If the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has eccentricity $e = \frac{5}{4}$ and the length of the latus rectum equal to $9$,then $ab$ is equal to

Similar Questions

The centre of the hyperbola $9x^{2} - 36x - 16y^{2} + 96y - 252 = 0$ is:

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $9 y^{2}-4 x^{2}=36$.

The equation of the tangent to the conic $x^2 - y^2 - 8x + 2y + 11 = 0$ at the point $(2, 1)$ is:

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$,then the equation of the tangent at one of the points of contact is

Vedclass Test Series

Exam Paper Generator

Online Exam Module