The distance between the foci of a hyperbola | vedclass

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

Question

(D) Given,the distance between the foci is $2ae = 16$. Since the eccentricity $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which implies $a = \frac{8}{\

Vedclass · Accepted Answer

$x^2-y^2=32$

Vedclass · Answer

(D) Given,the distance between the foci is $2ae = 16$. Since the eccentricity $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which implies $a = \frac{8}{\sqrt{2}} = 4\sqrt{2}$. For a hyperbola,$b^2 = a^2(e^2 - 1)$. Substituting the values,$b^2 = (4\sqrt{2})^2 ((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$. The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Substituting $a^2 = 32$ and $b^2 = 32$,we get $\frac{x^2}{32} - \frac{y^2}{32} = 1$,which simplifies to $x^2 - y^2 = 32$.

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

Similar Questions

If $L_1=0$ and $L_2=0$ are the asymptotes of the hyperbola $9x^2-4y^2+36x+8y-4=0$,then the product of the perpendicular distances from the point $(1,1)$ to the lines $L_1=0$ and $L_2=0$ is

Let $16 x^{2}-3 y^{2}-32 x-12 y=44$ represent a hyperbola. Then,

The value of $m$,for which the line $y = mx + \frac{25\sqrt{3}}{3}$ is a normal to the conic $\frac{x^2}{16} - \frac{y^2}{9} = 1$,is

If the perimeter of a triangle is $20$ and two of its vertices are $(-5, 0)$ and $(6, 0)$,then the locus of the third vertex is:

$A$ hyperbola with centre at $(0,0)$ has its transverse axis along the $X$-axis,and its length is $12$. If $(8,2)$ is a point on the hyperbola,then its eccentricity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module