On the ellipse frac{x^{2}}{8}+frac{y^{2}}{4}= | vedclass

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

Question

(B) The equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$. Here $a^{2}=8$ and $b^{2}=4$.The eccentricity $e$ is given by $e = \sqrt{1 - \

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$. Here $a^{2}=8$ and $b^{2}=4$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{4}{8}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.The foci are $S(-ae, 0)$ and $S'(ae, 0)$,where $ae = \sqrt{8} \cdot \frac{1}{\sqrt{2}} = 2$. So,$S(-2, 0)$ and $S'(2, 0)$.The line perpendicular to $x+2y=0$ has the form $2x-y+k=0$,or $y=2x+k$.The condition for tangency $y=mx+c$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}+b^{2}$.Here $m=2$,so $c^{2} = 8(2^{2}) + 4 = 32+4 = 36$,which gives $c = \pm 6$.Since $P$ is in the second quadrant,the tangent must have a positive $y$-intercept,so $y=2x+6$.The point of tangency $P(x_{1}, y_{1})$ is given by $(-\frac{a^{2}m}{c}, \frac{b^{2}}{c}) = (-\frac{8 \cdot 2}{6}, \frac{4}{6}) = (-\frac{8}{3}, \frac{2}{3})$.The area $A$ of $\Delta SPS'$ with base $SS'=4$ and height $|y_{1}| = \frac{2}{3}$ is $A = \frac{1}{2} \cdot 4 \cdot \frac{2}{3} = \frac{4}{3}$.Finally,$(5-e^{2}) \cdot A = (5 - \frac{1}{2}) \cdot \frac{4}{3} = \frac{9}{2} \cdot \frac{4}{3} = 6$.

Similar Questions

If $x+\sqrt{3} y=3$ is the tangent to the ellipse $2 x^2+3 y^2=k$ at a point $P$,then the equation of the normal to this ellipse at $P$ is

If $\alpha, \beta$ are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^2+4y^2=4$,then $\sqrt{3} \cos \frac{\alpha+\beta}{2} =$

The product of the perpendicular distances drawn from the points $(3,0)$ and $(-3,0)$ to the tangent of the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$ at the point $\left(3, \frac{9}{2}\right)$ is:

If the line joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ is a focal chord,then one possible value of $\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2}$ is

The conic represented by $x = 2(\cos t + \sin t), y = 5(\cos t - \sin t)$ is .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module