Find the equation of the hyperbola satisfying | vedclass

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

Question

(A) Given,the foci are $(\pm 3 \sqrt{5}, 0)$. Since the foci lie on the $x$-axis,the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}} - \

Vedclass · Accepted Answer

$\frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$

Vedclass · Answer

(A) Given,the foci are $(\pm 3 \sqrt{5}, 0)$. Since the foci lie on the $x$-axis,the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Here,$c = 3 \sqrt{5}$,so $c^{2} = (3 \sqrt{5})^{2} = 9 	imes 5 = 45$.The length of the latus rectum is given as $8$,so $\frac{2b^{2}}{a} = 8$,which implies $b^{2} = 4a$.We know the relation $c^{2} = a^{2} + b^{2}$. Substituting the values,we get $a^{2} + 4a = 45$,or $a^{2} + 4a - 45 = 0$.Factoring the quadratic equation: $(a + 9)(a - 5) = 0$. Since $a > 0$,we have $a = 5$.Then $b^{2} = 4a = 4 	imes 5 = 20$.Substituting $a^{2} = 25$ and $b^{2} = 20$ into the standard equation,we get $\frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$.

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 3 \sqrt{5}, 0)$,the latus rectum is of length $8$.

Similar Questions

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 product of the lengths of the perpendiculars from any point on the hyperbola $16x^2 - 25y^2 = 400$ to its asymptotes is $p$ and the angle between the two asymptotes is $\theta$,then $p \tan \frac{\theta}{2} =$

The equation of a line passing through the centre of a rectangular hyperbola is $x - y - 1 = 0$. If one of the asymptotes is $3x - 4y - 6 = 0$,the equation of the other asymptote is:

The reciprocal of the eccentricity of a rectangular hyperbola is

Find the equation of the hyperbola satisfying the given conditions: Vertices $(0, \pm 3)$,foci $(0, \pm 5)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module