Question

The equation of the normal to the curve $y^2=x^3$ at the point whose abscissa is $8$ is -

• A. $x\pm \sqrt{2}y=104$
• B. $x\pm 3\sqrt{2}y=104$
• C. $3\sqrt{2}x\pm y=104$
• D. None of these

Answer 1

Answer: (B)

$x\pm 3\sqrt{2}y=104$

Solving the curve at $x=8=2^3$
$y^2=x^3=2^9$
$\Rightarrow y=\pm 2^4\sqrt{2}=\pm 16\sqrt{2}$
So the points are $(8,\pm 16\sqrt{2})$
Now the curve is, $y^2=x^3$
$\Rightarrow 2y\times \frac{dy}{dx}=3x^2$
$\Rightarrow \left(\frac{dy}{dx}\right)=\frac{3x^2}{2y}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(8,\pm 16\sqrt{2})}=\frac{3\times 64}{\pm 2\times 16\sqrt{2}}=\pm 3\sqrt{2}$
Thus the slope of normal is $=\left(\frac{dy}{dx}\right)=\mp \frac{1}{3\sqrt{2}}$
Therefore, equation of normal is,
$(y\mp 16\sqrt{2})=\mp \frac{1}{3\sqrt{2}}(x-8)$
$\Rightarrow \mp 3\sqrt{2}y\pm 96=(x-8)$
$\Rightarrow x\pm 3\sqrt{2}y=104$
Hence, option 'B' is correct.