Let A be a vertex of the ellipse S equiv frac | vedclass

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

Question

(C) The ellipse $S$ is $\frac{x^2}{4}+\frac{y^2}{9}=1$. Its vertices are $(0, \pm 3)$. Let $A = (0, 3)$.For the ellipse $S^{\prime} \equiv \frac{x^2}{

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) The ellipse $S$ is $\frac{x^2}{4}+\frac{y^2}{9}=1$. Its vertices are $(0, \pm 3)$. Let $A = (0, 3)$.For the ellipse $S^{\prime} \equiv \frac{x^2}{9}+\frac{y^2}{4}=1$,the major axis is along the $x$-axis,$a^2=9, b^2=4$. The eccentricity $e = \sqrt{1-\frac{4}{9}} = \frac{\sqrt{5}}{3}$. The foci are $(\pm ae, 0) = (\pm \sqrt{5}, 0)$.Let $F = (\sqrt{5}, 0)$. Point $P$ divides $\overline{OF}$ in the ratio $2:1$. Thus,$P = \left(\frac{2(\sqrt{5})+1(0)}{2+1}, 0ight) = \left(\frac{2\sqrt{5}}{3}, 0ight)$.The line passing through $A(0, 3)$ and $P\left(\frac{2\sqrt{5}}{3}, 0ight)$ has the equation $y - 3 = \frac{0-3}{\frac{2\sqrt{5}}{3}-0}(x-0)$,which simplifies to $y = -\frac{9}{2\sqrt{5}}x + 3$.Substituting $y = 3 - \frac{9}{2\sqrt{5}}x$ into $\frac{x^2}{4} + \frac{y^2}{9} = 1$,we get $\frac{x^2}{4} + \frac{(3 - \frac{9}{2\sqrt{5}}x)^2}{9} = 1$.$\frac{x^2}{4} + \frac{9(1 - \frac{3}{2\sqrt{5}}x)^2}{9} = 1$ $\Rightarrow \frac{x^2}{4} + 1 - \frac{3}{\sqrt{5}}x + \frac{9}{20}x^2 = 1$.$\frac{5x^2 + 9x^2}{20} = \frac{3}{\sqrt{5}}x$ $\Rightarrow \frac{14x^2}{20} = \frac{3}{\sqrt{5}}x$ $\Rightarrow x = 0$ or $x = \frac{3}{\sqrt{5}} 	imes \frac{20}{14} = \frac{30}{7\sqrt{5}}$.For $x = \frac{30}{7\sqrt{5}}$,$y = 3 - \frac{9}{2\sqrt{5}}(\frac{30}{7\sqrt{5}}) = 3 - \frac{270}{70} = 3 - \frac{27}{7} = -\frac{6}{7}$.The chord length is $\sqrt{(\frac{30}{7\sqrt{5}} - 0)^2 + (-\frac{6}{7} - 3)^2} = \sqrt{\frac{900}{49 	imes 5} + (-\frac{27}{7})^2} = \sqrt{\frac{180}{49} + \frac{729}{49}} = \sqrt{\frac{909}{49}} = \frac{3\sqrt{101}}{7}$.Thus,$k=7$.

Similar Questions

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval:

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 eccentric angle of a point on the ellipse $x^2+3y^2=6$ lying at a distance of $2$ units from its centre is

The values that $m$ can take,so that the straight line $y=4x+m$ touches the curve $x^2+4y^2=4$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module