Find the equation of normal to parabola `y=x^(2)-3x-4`
(a) at point (3,-4)
(b) having slope 5.

Answer

(a) `3x-y-13=0`, (b) `5x-y-20=0`

Explanation

We have parabola
`y=x^(2)-3x-4`
Differentiating w.r.t. x, we get
`(dy)/(dx)=2x-3`
(a) `((dy)/(dx))_((2","-4))=3`
So, equation of normal is
`4+4=3(x-3)`
`or" "3x-y-13=0`
(b) `(dy)/(dx)=2x-3=5`
`:." "x=4`
So, y=16-12-4=0.
Therefore, equation of normal is
`y-0=5(x-4)`
`or" "5x-y-20=0`