Find the equation of the hyperbola whose foci | vedclass

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

Question

(A) Let the center of the hyperbola be $(\alpha, \beta)$.Since the foci lie on the line $y=5$,the transverse axis is horizontal.Therefore,the equation

Vedclass · Accepted Answer

$\frac{(x-1)^2}{16} - \frac{(y-5)^2}{9} = 1$

Vedclass · Answer

(A) Let the center of the hyperbola be $(\alpha, \beta)$.Since the foci lie on the line $y=5$,the transverse axis is horizontal.Therefore,the equation is of the form $\frac{(x-\alpha)^2}{a^2} - \frac{(y-\beta)^2}{b^2} = 1$.The center is the midpoint of the foci: $\alpha = \frac{6+(-4)}{2} = 1$ and $\beta = 5$.So,the center is $(1, 5)$.The distance between the foci is $2ae = 6 - (-4) = 10$.Given $e = 5/4$,we have $2a(5/4) = 10$,which implies $a = 4$.Using the relation $b^2 = a^2(e^2 - 1)$,we get $b^2 = 16(\frac{25}{16} - 1) = 16(\frac{9}{16}) = 9$.Substituting these values into the standard equation,we get $\frac{(x-1)^2}{16} - \frac{(y-5)^2}{9} = 1$.

Find the equation of the hyperbola whose foci are $(6, 5)$ and $(-4, 5)$ and eccentricity is $5/4$.

Similar Questions

The equation of one of the tangents drawn from the point $(0, 1)$ to the hyperbola $45x^2 - 4y^2 = 5$ is

If the line $y=mx+c$ is a common tangent to the hyperbola $\frac{x^{2}}{100}-\frac{y^{2}}{64}=1$ and the circle $x^{2}+y^{2}=36,$ then which one of the following is true?

The normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ at the point $(8, 3\sqrt{3})$ on it passes through the point

The equation $13[{(x - 1)^2} + {(y - 2)^2}] = 3{(2x + 3y - 2)^2}$ represents

If $lx + my = 1$ is a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then $a^2 m^2 - b^2 l^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module