B is an extremity of the minor axis of an ell | vedclass

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

Question

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The coordinates of the foci are $S(ae, 0)$ and $S^{\prime

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$.The extremity of the minor axis is $B(0, b)$.The slope of $SB$ is $m_1 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.The slope of $S^{\prime}B$ is $m_2 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.Since $\angle SBS^{\prime} = 90^{\circ}$,the product of the slopes is $-1$,so $m_1 	imes m_2 = -1$.$(-\frac{b}{ae}) 	imes (\frac{b}{ae}) = -1$.$\frac{b^2}{a^2 e^2} = 1 \Rightarrow b^2 = a^2 e^2$.Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2 e^2$.$1 - e^2 = e^2$ $\Rightarrow 2e^2 = 1$ $\Rightarrow e^2 = \frac{1}{2}$.Since eccentricity $e > 0$,we have $e = \frac{1}{\sqrt{2}}$.

$B$ is an extremity of the minor axis of an ellipse whose foci are $S$ and $S^{\prime}$. If $\angle SBS^{\prime}$ is a right angle,then the eccentricity of the ellipse is

Similar Questions

Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci,then the square of the sum of the lengths of its minor and major axes is equal to $...........$.

If the normal at any point $P$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ cuts the major and minor axes in $G$ and $g$ respectively,and $C$ is the centre of the ellipse,then:

If the distance between the foci of an ellipse is $6$ and the length of the minor axis is $8$,then the eccentricity is

The length of the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$,whose mid-point is $\left(1, \frac{1}{2}\right)$,is:

Statement $I$: The equation of the directrix of the ellipse $4x^2+y^2-8x-4y+4=0$ is $3y=6-4\sqrt{3}$.
Statement $II$: The equation of the latus rectum of the ellipse $x^2+4y^2-4x-8y+4=0$ is $y=2+\sqrt{3}$.
Which of the above statement$(s)$ is (are) true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module