Question

The equation of the chord of the parabola $y^2=8x$ which is bisected at the point $(2,-3)$, is.

• A. $4x+3y+1=0$
• B. $3x+4y-1-0$
• C. $4x-3y-1=0$
• D. $3x-4y+1=0$

Answer 1

Answer: (A)

$4x+3y+1=0$

$y^2=8x$ bisected at $(2,-3)$

$S=y^2-8x=0$

$\therefore S_1=Sy$

$\Rightarrow y{y}^{\prime }-8\left(\frac{x+x^{\prime }}{2}\right)=y^2-8x^{\prime }$

$\Rightarrow y(-3)-8\left(\frac{x+2}{2}\right)={(-3)}^2-8\left(2\right)$

$\Rightarrow$ $-3y-4x-8=9-16$

$\Rightarrow$ $-3y-4x-8+7=0$

$\Rightarrow$ $-4x-3y-1=0$

$4x+3y+1=0$ is eqn. of chord.