Finding the cutting point of a normal to a curve!

I have been trying to solve this by using simultaneous equations but have failed so far.

- Find the point where normal to (2,4) to y = x^2 cuts the curve again.

(1) Since the gradient of y = x^2 = 2x, gradient of the tangent to the curve should be 4. Therefore, the gradient of the normal should -1/4.

Then, y - 4 = -1/4 (x - 2) => y = -1/4x + 9/2

So I think equation of the normal should be: y = -1/4x + 9/2.. Next solving the equations simulatenously, I am left with:

x^2 + 1/4x - 9/2 = 0

But I cannot find the roots of this which would help me reach the final solution by finding two points where these two equations meet. Is there any better way of solving this equation and what am I doing wrong?

P.S. I have just begun on the differentiation.

Cutting point to normal of a curve

Quote:

Originally Posted by

**struck** I have been trying to solve this by using simultaneous equations but have failed so far.

- Find the point where normal to (2,4) to y = x^2 cuts the curve again.

(1) Since the gradient of y = x^2 = 2x, gradient of the tangent to the curve should be 4. Therefore, the gradient of the normal should -1/4.

Then, y - 4 = -1/4 (x - 2) => y = -1/4x + 9/2

So I think equation of the normal should be: y = -1/4x + 9/2.. Next solving the equations simulatenously, I am left with:

x^2 + 1/4x - 9/2 = 0

But I cannot find the roots of this which would help me reach the final solution by finding two points where these two equations meet. Is there any better way of solving this equation and what am I doing wrong?

P.S. I have just begun on the differentiation.

You are correct so far.

Now multiply both sides by 4 to remove the fractions

4x^2 + x - 9 = 0

Factorise: (4x + 9)(x - 2) = 0

Roots: x = -9/4, x = 2

When x = -9/4, y = 81/16

cutting point (-9/4, 81/16)