Question

If the line $4x+3y+1=0$ meets the parabola $y^2=8x$ then mid point of the chord is

• A. (-1, 1)
• B. (2,-3)
• C. (3,-3)
• D. (5,-7)

Answer 1

The correct option is B (2,-3)

$\begin{array}{c}LetP\left(x_1,y_1\right)andQ\left(x_2,y_2\right)\\ Now,\\ 3y=-4x-1\\ y=\frac{-4x-1}{3}\\ 16x^2+1+8x=72x\\ 16x^2-64x+1=0\\ x_1+x_1=\frac{64}{16}=4\\ Now,M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ Then\\ y_1+y_2=\frac{-4x_1-1}{3}-\frac{4x_2-1}{3}\\ =\frac{-4\left(x_1+x_2\right)-1}{3}\\ =\frac{-18}{3}=-6\\ So,\\ M=\left(2,-3\right)\end{array}$