Find the equation of tangent & normal to the curve $y^4=a{x}^3\text{ at }(a,a)$

Given $P(a, a)$

curve $y^4=a{x}^3$

diff. w.r.t ‘$x$’ 

$4y^3\frac{dy}{dx}=a3x^2 \frac{dy}{dx}=\frac{3a{x}^2}{4y^3}$

slope $\left(m\right)={\left(\frac{dy}{dx}\right)}_{P(a,a)}$

$=\frac{3a\cdot a^2}{4a^3}; m\cdot =\frac{3}{4}$

1) Equationof tangent

$y-y_1=m\left(x,x_1\right)$

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

$4y-4a=3x-3a$

$3x-4y+a=0$

2) Equation of normal

$y-y_1=\frac{-1}{m}\left(x-x_1\right)$

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

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

$4x+3y-7a=0$