Let A, A^{prime} be the end points of the maj | vedclass

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

Question

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The coordinates are $A(a, 0)$,$B(0, b)$,and $B^{\prime}(0, -b)$.Given

Vedclass · Accepted Answer

$2 	an ^{-1}(\sqrt{2})$

Vedclass · Answer

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The coordinates are $A(a, 0)$,$B(0, b)$,and $B^{\prime}(0, -b)$.Given $\angle BAB^{\prime} = 60^{\circ}$,the line $AB$ makes an angle of $30^{\circ}$ with the major axis $AA^{\prime}$.In $	riangle OAB$,$	an 30^{\circ} = \frac{OB}{OA} = \frac{b}{a}$.Thus,$\frac{b}{a} = \frac{1}{\sqrt{3}}$,which implies $a = \sqrt{3}b$.The distance of the focus $S$ from the center $O$ is $c = \sqrt{a^2 - b^2} = \sqrt{3b^2 - b^2} = \sqrt{2}b$.Let $\angle SBS^{\prime} = 2	heta$,where $	heta = \angle OBS$. In $	riangle OBS$,$	an 	heta = \frac{OS}{OB} = \frac{c}{b} = \frac{\sqrt{2}b}{b} = \sqrt{2}$.Therefore,$\angle SBS^{\prime} = 2	heta = 2 	an^{-1}(\sqrt{2})$.

Let $A, A^{\prime}$ be the end points of the major axis,$S, S^{\prime}$ be the foci,and $B, B^{\prime}$ be the end points of the minor axis of an ellipse $E$. If $\angle BAB^{\prime}=60^{\circ}$,then find $\angle SBS^{\prime}$.

Similar Questions

If the area of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{\lambda^{2}}=1$ is $20 \pi$ sq units,then $\lambda$ is

The value of $k$ such that the line $y=2x+k$ touches the ellipse $3x^2+5y^2=15$ is

The area enclosed by the ellipse $4x^2 + 9y^2 = 1$ is . . . . . . sq. units.

If tangents are drawn from any point on the circle $x^2+y^2=25$ to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$,then the angle between the tangents is

For the ellipse $\frac{x^2}{64} + \frac{y^2}{28} = 1$,the eccentricity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module