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

For the hyperbola $x^2 \sec^2 \theta - y^2 \csc^2 \theta = 1$,which of the following remains constant when $\theta$ varies?

The product of the lengths of the perpendiculars drawn from the foci to any tangent to the hyperbola $x^{2} - \frac{y^{2}}{4} = 1$ is:

The locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}k = 0$ and $\sqrt{3}kx + ky - 4\sqrt{3} = 0$ for different values of $k$ is

If the variable line $y = kx + 2h$ is tangent to an ellipse $2x^2 + 3y^2 = 6$,the locus of $P(h, k)$ is a conic $C$ whose eccentricity equals

If the equation $x+y+n=0$ represents a normal to the hyperbola $\frac{x^2}{6}-\frac{y^2}{2}=1$,then $n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module