Question

If the tangent at the point P(2, 4) to the parabola $y^2=8x$ meets the parabola $y^2=8x+5$ at Q and R,
then the midpoint of QR is

• A. (2, 4)
• B. (4, 2)
• C. (7, 9)
• D. (3, 3)

Answer 1

The correct option is A

(2, 4)

For the parabola $y^2=8x,$ the point P(2, 4) corresponds to t =1.
slope of targent at $a{t}^2,2at$ is $\frac{1}{t}$
Hence the tangent to the parabola $y^2=8x$ at (2, 4) is
$y-4=\frac{1}{1}(x-2)\Rightarrow x-y+2=0\Rightarrow y=x+2y=x+2intersectstheparabola{y}^2=8x+5\Rightarrow (x+2{)}^2=8x+5\Rightarrow x^2+4x+4=8x+5\Rightarrow x^2-4x-1=0Solvingforx,x=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}$
corresponding y coordinates are
$x+2\Rightarrow 4\pm \sqrt{3}\Rightarrow Q=(2+\sqrt{3},4+\sqrt{3})\Rightarrow R=(2-\sqrt{3},4-\sqrt{3})$
The midpoint of QR =(2, 4)