Question

The mid-point of the line joining the common points of the line $2x-3y+8=0$ and $y^2=8x$ is

• A. $(3,2)$
• B. $(5,6)$
• C. $(4,-1)$
• D. $(2,-3)$

Answer 1

The correct option is B $(5,6)$

Let $(x_1,y_1)$ be the mid-point of the line joining the common points of the given line and the given parabola.

Then, the equation of line is
$y{y}_1-4(x+x_1)=y_1^2-8x_1$
$\Rightarrow 4x=y{y}_1+y_1^2-4x_1=0$
This line and $2x-3y+8=0$ represents the same line.
Therefore, $\frac{4}{2}=\frac{-y_1}{-3}=\frac{y_1^2-4x_1}{8}$
$\Rightarrow y_1=6,y_1^2-4x_1=16$
$\Rightarrow y_1=6,x_1=\frac{36-16}{4}$
$\Rightarrow y_1=6,x_1=5$
Thus the required point is $(5,6)$.