Q95 / 160 · TS EAMCET 2013 · Mathematics

The mid-point of a chord of the ellipse $x^2+4 y^2-2 x+20 y=0 \quad$ is $(2,-4)$. The equation of the chord is

  • Correct answer
    A
    $x-6 y=26$
  • B
    $x+6 y=26$
  • C
    $6 x-y=26$
  • D
    $6 x+y=26$

Solution

Equation of chord at $(2,-4)$ is $$ \begin{aligned} & T=S^{\prime} \\ & \Rightarrow \quad 2 x+4 y(-4)-(x+2)+10(y-4) \\ & =(2)^2+4(-4)^2+2(2)+20(-4) \\ & \Rightarrow \quad 2 x-16 y-x-2+10 y-40 \\ & =4+64-4-80 \\ & \Rightarrow \quad x-6 y=42-16 \Rightarrow 26 \\ & \end{aligned} $$

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

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