Question

Equations of normals to the curve $y^2=x^3$ at the point whose abscissa x is 8 be $x\pm k\sqrt{m}y=2p$. Find $\frac{p}{m}+k$

Answer 1

$y^2=x^3$
At $x=8$, $y=(8^3{)}^{1/2}$
$=2^{9/2}$
Hence the point is
$(8,2^{9/2})$.
The slope of the normal is
$\frac{-dx}{dy}$
Now $2y.dy=3x^{2}dx$
$\frac{dx}{dy}=\frac{2y}{3x^2}$
$\frac{-dx}{dy}=\frac{-2y}{3x^2}$
Hence the slope of the normal is
$m=\frac{-2\times 2^{9/2}}{3\times 64}$
$=\frac{-1}{3}\left(2^{11/2-6}\right)$
$=\frac{-1}{3\sqrt{2}}$
$=\frac{-\sqrt{2}}{6}$
Hence the equation of the normal is
$y-2^{9/2}=(x-8)\left(\frac{-\sqrt{2}}{6}\right)$
$y-16\sqrt{2}=(x-8)\left(\frac{-\sqrt{2}}{6}\right)$
$6y-96\sqrt{2}=-\sqrt{2}x+8\sqrt{2}$
$\sqrt{2}x+6y=104\sqrt{2}$
$x+3\sqrt{2}y=104$
Hence
$k=3,m=2$ and
$2p=104$
$p=52$
Hence
$\frac{p}{m}+k$
$=\frac{52}{2}+3$
$=26+3$
$=29$