If theta is the angle subtended by a latus re | vedclass

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

Question

(C) The eccentricity is given by $e = \frac{2}{\sqrt{7}-\sqrt{3}}$. Rationalizing the denominator,we get $e = \frac{2(\sqrt{7}+\sqrt{3})}{7-3} = \frac

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) The eccentricity is given by $e = \frac{2}{\sqrt{7}-\sqrt{3}}$. Rationalizing the denominator,we get $e = \frac{2(\sqrt{7}+\sqrt{3})}{7-3} = \frac{2(\sqrt{7}+\sqrt{3})}{4} = \frac{\sqrt{7}+\sqrt{3}}{2}$.For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the coordinates of the ends of the latus rectum are $(ae, \pm \frac{b^2}{a})$.The angle $	heta$ subtended by the latus rectum at the centre $(0,0)$ is given by $	an(\frac{	heta}{2}) = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.Using $b^2 = a^2(e^2-1)$,we have $	an(\frac{	heta}{2}) = \frac{a^2(e^2-1)}{a^2e} = \frac{e^2-1}{e}$.Given $e = \frac{\sqrt{7}+\sqrt{3}}{2}$,then $e^2 = \frac{7+3+2\sqrt{21}}{4} = \frac{10+2\sqrt{21}}{4} = \frac{5+\sqrt{21}}{2}$.So,$	an(\frac{	heta}{2}) = \frac{\frac{5+\sqrt{21}}{2} - 1}{\frac{\sqrt{7}+\sqrt{3}}{2}} = \frac{3+\sqrt{21}}{\sqrt{7}+\sqrt{3}} = \frac{\sqrt{3}(\sqrt{3}+\sqrt{7})}{\sqrt{7}+\sqrt{3}} = \sqrt{3}$.Thus,$\frac{	heta}{2} = 60^{\circ}$,which means $	heta = 120^{\circ}$.Therefore,$\sin 	heta = \sin 120^{\circ} = \frac{\sqrt{3}}{2}$.

If $\theta$ is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity $e = \frac{2}{\sqrt{7}-\sqrt{3}}$,then $\sin \theta = $

Similar Questions

The tangent drawn at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\frac{x^2}{4}-\frac{y^2}{5}=1$ meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. If $O$ is the origin,then $(OA)^2-(OB)^2=$

The equation of the tangents to the conic $3x^2 - y^2 = 3$ perpendicular to the line $x + 3y = 2$ is

If the eccentricities of two conics $S$ and $S'$ are $e$ and $e'$ respectively,such that $e^2 + e'^2 = 3$,then both $S$ and $S'$ are:

Let the product of the focal distances of the point $P(4, 2\sqrt{3})$ on the hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $32$. Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be $q$. Then $p^2 + q^2$ is equal to ......

The eccentricity of the hyperbola $4x^2 - 9y^2 = 16$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module