Question

How do you find the equations for the normal line to `y=x^2` through (2,4)?

Answer 1

`y=-1/4x+9/2`

Explanation:

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent, so the product of their gradients is `-1`

so If `y=x^2` then differentiating wrt `x` gives us:

`dy/dx = 2x`

When `x=2 => y=2^2=4` (so `(2,4)` lies on the curve)
and `dy/dx=(2)2=4`

So the normal we seek passes through `(2,4)` ad has gradient `-1/4` so using `y-y_1=m(x-x_1)` the equation we seek is;

`y-4=-1/4(x-2)`
`:. y-4=-1/4x+1/2`
`:. y=-1/4x+9/2`

We can confirm this graphically: