Question

Find the coordinates of point on the curve $y^2=8x,$ which is at minimum distance from the line $x+y=-2.$

• A. $(-2,-4)$
• B. $(2,4)$
• C. $(2,-4)$
• D. $(-2,4)$

Answer 1

Answer: (C)

$(2,-4)$

Let the parametric points on the curve be $2t^2,4t$
Now perpendicular distance of the point from the line will be
$S=\frac{4t+2t^2+2}{\sqrt{2}}$
Now
$\frac{dS}{dt}$
$=\frac{1}{\sqrt{2}}[4+4t]=0$
Or
$t=-1$.
Hence we get the point as
$(2,-4)$.