If the latus rectum of a hyperbola frac{x^2}{ | vedclass

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

Question

(D) The foci of the hyperbola are $S(ae, 0)$ and $S'(-ae, 0)$.The endpoints of the latus rectum passing through $S$ are $A(ae, \frac{b^2}{a})$ and $B(

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) The foci of the hyperbola are $S(ae, 0)$ and $S'(-ae, 0)$.The endpoints of the latus rectum passing through $S$ are $A(ae, \frac{b^2}{a})$ and $B(ae, -\frac{b^2}{a})$.The angle subtended by $AB$ at $S'$ is $60^{\circ}$.Since the triangle $	riangle AS'B$ is isosceles,the line $S'S$ bisects the angle $\angle AS'B$,so $\angle AS'S = 30^{\circ}$.The slope of $S'A$ is $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$.Thus,$\frac{b^2/a}{ae - (-ae)} = \frac{b^2}{2a^2e} = \frac{1}{\sqrt{3}}$.This gives $\frac{b^2}{a^2} = \frac{2e}{\sqrt{3}}$.Using the relation $b^2 = a^2(e^2 - 1)$,we have $\frac{b^2}{a^2} = e^2 - 1$.Equating the two expressions: $e^2 - 1 = \frac{2e}{\sqrt{3}}$,which simplifies to $\sqrt{3}e^2 - 2e - \sqrt{3} = 0$.Solving the quadratic equation: $(\sqrt{3}e + 1)(e - \sqrt{3}) = 0$.Since the eccentricity $e > 1$,we get $e = \sqrt{3}$.

If the latus rectum of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends an angle of $60^{\circ}$ at the other focus,then the eccentricity of the hyperbola is

Similar Questions

The locus of the point of intersection of the lines $(\sqrt{3})kx + ky - 4\sqrt{3} = 0$ and $\sqrt{3}x - y - 4\sqrt{3}k = 0$ is a conic,whose eccentricity is .............

Let $S$ be the focus of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ lying on the positive $X$-axis and $P(5, y_1)$ be a point on the hyperbola. Then $SP =$

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm \sqrt{10})$,passing through $(2, 3)$.

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$,be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

For the hyperbola $x^2-y^2-4x+2y+c=0$,if the focus is $S(2+2\sqrt{2}, k)$ and the directrix that is adjacent to $S$ is $x=2+\sqrt{2}$,then $c=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module