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=a

Answer 1

The correct option is C 4x+3y=7a

The equation of the curve is,

$y^4=a{x}^3$

Differentiate w.r.t x, we get,

$4y^3\frac{dy}{dx}=3a{x}^2$

$\therefore \frac{dy}{dx}=\frac{3a{x}^2}{4y^3}$

$\therefore {\frac{dy}{dx}}_{(a,a)}=\frac{3a{\left(a\right)}^2}{4{\left(a\right)}^3}$

$\therefore {\frac{dy}{dx}}_{(a,a)}=\frac{3{\left(a\right)}^3}{4{\left(a\right)}^3}$

$\therefore {\frac{dy}{dx}}_{(a,a)}=\frac{3}{4}$

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,

$y-y_1=m_1(x-x_1)$

$\therefore y-a=\frac{-4}{3}(x-a)$

$\therefore 3(y-a)=-4(x-a)$

$\therefore 3y-3a=-4x+4a$

$\therefore 4x+3y=7a$