Question

The coordinates of the point $P$ on the curve $y^2=2x^3$ the tangent at which is perpendicular to the line $4x-3y+2=0$, are given by

• A. $(2,4)$
• B. $(1,\sqrt{2})$
• C. $(1/2,-1/2)$
• D. $(1/8,-1/16)$

Answer 1

The correct option is D $(1/8,-1/16)$

$4x-3y+2=0$
Or
$3y=4x+2$
Or
$y=\frac{4}{3}x+\frac{2}{3}$.
Hence slope of the tangent will be
$=\frac{-3}{4}$
Now
$\frac{dy}{dx}=\frac{-3}{4}$
$y=\pm \sqrt{2}.x^{\frac{3}{2}}$
Hence
$\frac{dy}{dx}$
$=\pm \sqrt{2}.\frac{3}{2}.x^{\frac{1}{2}}$
$=\frac{-3}{4}$
Or
$-\sqrt{2}.\frac{3}{2}.x^{\frac{1}{2}}=\frac{-3}{4}$
Or
$2\sqrt{2}.x^{\frac{1}{2}}=1$
Or
$x=\frac{1}{8}$.
Hence
$y=-\sqrt{2}.x^{\frac{3}{2}}$
$=-\sqrt{2}.\frac{1}{16\sqrt{2}}$
$=-\frac{1}{16}$
Hence the co-ordinates are $(\frac{1}{8},\frac{-1}{16})$