# Finding the cutting point of a normal to a curve!

• Oct 21st 2008, 02:07 AM
struck
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.
• Oct 21st 2008, 02:14 AM
earboth
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.

First: All your considerations and calculations are correct! (Clapping)

You can use the quadratic formula to solve this equation for x.

Since you already know one solution (x = 2 must be a solution of this equation because you calculate all intercepts!) you can try to factor out (x-2):

$\displaystyle x^2 + 1/4x - 9/2 = x^2 - 2x + \dfrac94x - \dfrac92 = x(x-2)+\dfrac94(x-2) = (x-2)\left(x+\dfrac94\right)$

• Oct 21st 2008, 02:40 AM
struck
Thanks :) . But are you sure I am doing it correctly? Because the quadratic formula or completing the square isn't helping me in finding the roots.

Edit: Ohh, but factoring it out like you did gets me the solutions. Weird :S .. Why can't the quadratic formula be applied directly? Or perhaps I did something wrong while applying the formula though I tried it twice.

Edit: but 2, and -9/4 isn't the answer so I did something wrong :(
• Oct 21st 2008, 07:36 AM
earboth
Quote:

Originally Posted by struck
Thanks :) . But are you sure I am doing it correctly? Yes! Because the quadratic formula or completing the square isn't helping me in finding the roots.

Why can't the quadratic formula be applied directly? Or perhaps I did something wrong while applying the formula
...

The quadratic equation $\displaystyle ax^2 + bx + c = 0$ has the solutions

$\displaystyle x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\displaystyle x = \dfrac{-\frac14 \pm \sqrt{\frac1{16}-4 \cdot 1 \cdot \left(-\frac92\right)}}{2 \cdot 1} = \dfrac{-\frac14 \pm \sqrt{\frac1{16}+\frac{288}{16}}}{2 \cdot 1}$

and you'll get the values I've posted before.

By the way: The second point is of course $\displaystyle Q\left(-\dfrac94~,~\dfrac{81}{16}\right)$
• Dec 23rd 2008, 05:59 AM
M_O'Loughlin
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)