Q93 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

$S=(-1,1)$ is the focus, $2 x-3 y+1=0$ is the directrix corresponding to $S$ and $\frac{1}{2}$ is the eccentricity of an ellipse. If $(a, b)$ is the centre of the ellipse, then $3 a+2 b=$

  • A
    $\frac{30}{13}$
  • B
    $\frac{4}{13}$
  • Correct answer
    C
    -1
  • D
    0

Solution

Given that focus $=(-1,1)$ and directrix $2 x-3 y+1=0$ and $e=\frac{1}{2}$ $\therefore$ Equation of ellipse $(x+1)^2+(y-1)^2=\left(\frac{1}{2}\right)^2\left(\frac{2 x-3 y+1}{\sqrt{4+9}}\right)^2$ $\begin{aligned} & \Rightarrow 48 x^2+43 y^2+100 x-98 y+103+12 x y=0 \\ & f(x, y)=48 x^2+43 y^2+100 x-98 y+103+12 x y\end{aligned}$ $\Rightarrow 24 x+3 y+25=0$..(i) $f_y=12 x+86 y-98=0$ $6 x+43 y-49=0$..(ii) On solving (i) and (ii), we get $\begin{aligned} & x=\frac{-661}{169 \times 3}=a \text { and } y=\frac{221}{169}=b \\ & \therefore 3 a+2 b=\frac{-661}{169}+\frac{442}{169}=-1\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions