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 vertices of the hyperbola $7x^2 - 49y^2 = 343$ are

Let $P(h, k)$ be the point of contact of the tangent to the hyperbola $5 x^2-7 y^2-35=0$ which is parallel to the line $\sqrt{2} x-y+\lambda=0$. If $P$ lies in the third quadrant,then $3 h^2-2 k=$

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

The equation of the line passing through the points $\left(ct_1, \frac{c}{t_1}\right)$ and $\left(ct_2, \frac{c}{t_2}\right)$ is

If the latus rectum through one of the foci of a hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtends a right angle at the farther vertex of the hyperbola,then $b^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module