The equation of the hyperbola referred to its | vedclass

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

Question

(B) The distance between the foci of a hyperbola is given by $2ae = 16$. Given $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which implies $a = \frac{8}{

Vedclass · Accepted Answer

$x^2 - y^2 = 32$

Vedclass · Answer

(B) The distance between the foci of a hyperbola is given by $2ae = 16$. Given $e = \sqrt{2}$,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,$b^2 = (4\sqrt{2})^2 ((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$. Thus,$b = 4\sqrt{2}$. The 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 equation of the hyperbola referred to its axes as axes of coordinate and whose distance between the foci is $16$ and eccentricity is $\sqrt{2}$ is:

Similar Questions

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

$A$ hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the point that lies on the tangent drawn to this hyperbola at $P$ is

The equations of the asymptotes of a hyperbola are $x+y+3=0$ and $2x-y+1=0$. If $(1,-2)$ is a point on this hyperbola,find the equation of its conjugate hyperbola.

The length of the transverse axis of the hyperbola $3x^2 - 4y^2 = 32$ is

The eccentricity of the hyperbola $5x^2 - 4y^2 + 20x + 8y = 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module