Question

the normal to the curve 5x^5-10x^3#x#2y#6 is equal to 0 at the point p(0,-3) is tangent to the curve at point

Answer 1

Step-by-step explanation:

Y=5×5+10×3+x+2y+6 let y=5×5/2+5×3-x/2-3

Answer 2

Answer:

The line is tangent at (1, – 1) and (– 1, – 5).

Step-by-step explanation:

Differentiating, 25x^4 – 30x^2 + 1 + 2y' = 0

At P(0, – 3), y' = –1/2

The normal at P is y + 3 = 2x

Eliminating y with the given equation x(x^2 – 1)^2 = 0

x = 0, 1, – 1

The line is tangent at (1, – 1) and (– 1, – 5).