Q96 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
$L_1^{\prime}$ is the end of a latus rectum of the ellipse $3 x^2+4 y^2$ $=12$ which is lying in the third quadrant. If the normal drawn at $L_1^{\prime}$ to this ellipse intersects the ellipse again at the point $P(a, b)$ then $a=$
- A$\frac{63}{38}$
- Correct answerB$\frac{11}{19}$
- C$-\frac{11}{19}$
- D$-\frac{63}{38}$
Solution
$3 x^2+4 y^2=12 \qquad ....\mathrm{(i)}$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow a^2=4, b^2=3, e=\frac{1}{2}$
End of latus rectum in third quadrant is $\left(-a e,-\frac{b^2}{a}\right)$
$\Rightarrow L_1^{\prime}=\left(-1, \frac{-3}{2}\right)$
Eq. of normal passing $L_1^{\prime}$ ellipse is $-4 x+2 y=1 \Rightarrow y=\frac{-1+4 x}{2}$ $\begin{aligned} & \text { From (i), } 3 x^2+(1+4 x)^2=12 \\ & \Rightarrow 19 x^2+8 x-11=0 \\ & \Rightarrow x=-1, \frac{11}{19} ; \text { Hence, } a=\frac{11}{19}\end{aligned}$
Eq. of normal passing $L_1^{\prime}$ ellipse is $-4 x+2 y=1 \Rightarrow y=\frac{-1+4 x}{2}$ $\begin{aligned} & \text { From (i), } 3 x^2+(1+4 x)^2=12 \\ & \Rightarrow 19 x^2+8 x-11=0 \\ & \Rightarrow x=-1, \frac{11}{19} ; \text { Hence, } a=\frac{11}{19}\end{aligned}$
Hence, the correct answer is option: 'B'.