Question

The equation of the normal to the curve $y^4=a{x}^3$ at (a, a) is

• A. $x+2y=3a$
• B. $3x-4y+a=0$
• C. $4x+3y=7a$
• D. $4x-3y=0$

Answer 1

Answer: (C)

The equation of the curve is,
$y^4=a{x}^3$
Differentiate w.r.t $x$, we get,
$4y^3\frac{dy}{dx}=3a{x}^2$
$\therefore \frac{dy}{dx}=\frac{3a{x}^2}{4y^3}$
$\therefore \frac{dy}{dx(a,a)}=\frac{3a(a{)}^2}{4(a{)}^3}$
$\therefore \frac{dy}{dx(a,a)}=\frac{3(a{)}^3}{4(a{)}^3}$
$\therefore \frac{dy}{dx(a,a)}=\frac{3}{4}$
Thus, slope of tangent is $\frac{3}{4}$
Let $m_1=$ slope of normal
As tangent and normal are perpendicular to each other, we can write,
$\frac{3}{4}\times m_1=-1$
$\therefore m_1=\frac{-4}{3}$
Thus, equation of normal passing through $(a,a)$ is,
$y-y_1=m_1\left(x-x_1\right)$
$\therefore y-a=\frac{-4}{3}(x-a)$
$\therefore 3(y-a)=-4(x-a)$
$\therefore 3y-3a=-4x+4a$
$\therefore 4x+3y=7a$