The difference of the focal distances of any | vedclass

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

Question

(A) 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 thi

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) 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 $a^2 = 16$,which implies $a = 4$.The difference of the focal distances of any point on a hyperbola is equal to the length of its transverse axis,which is $2a$.Therefore,the difference is $2 	imes 4 = 8$.

The difference of the focal distances of any point on the hyperbola $9x^2 - 16y^2 = 144$ is

Similar Questions

The equation of the directrices of the hyperbola $3x^{2}-3y^{2}-18x+12y+2=0$ is

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is

The locus of a point of intersection of two lines $x \sqrt{3}-y=k \sqrt{3}$ and $\sqrt{3} k x+k y=\sqrt{3}$,where $k \in R$,describes:

For the variable $t,$ the locus of the points of intersection of lines $x-2y=t$ and $x+2y=\frac{1}{t}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module