Question

Find the equation of the normal at the point (am^2 , am^ 3 ) for the curve ay^ 2 = x^ 3 .

Answer 1

HELLO DEAR,

GIVEN ;- curve is ay² = x³ at point (am² , am³).

now, differentiating curve with respect to x.

$\bold{2ay*dy/dx = 3x^2}$

$\bold{dy/dx = 3x^2/2ay}$

the slope of normal at point (am² , am³) is;

$\bold{dy/dx_{am^2 , am^3} = 3(am^2)^2/2a(am^3)}$

$\bold{-1/dy/dx_{am^2 , am^3} = -2/3m}$

now, the Equation of tangent at point (am² , am³) is;

y - am³ = (-2/3m)(x - am²)

3ym - 3am⁴ = -2x + 2am²

2x + 3ym = am²(2 + 3m²)

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.