Line intercepting curve

• Aug 27th 2010, 10:15 PM
caramelcake
Line intercepting curve
Quote:

A straight line of gradient $\displaystyle m$ is drawn through the point $\displaystyle A(2, -4)$ on the curve $\displaystyle y = x^2 - 4x$. Find, in terms of $\displaystyle m$, the coordinates of the point $\displaystyle B$ at which the line intercepts the curve again.
These are the steps I have done thus far:
Let $\displaystyle c$ be the $\displaystyle y$ - intercept
$\displaystyle -4 = 2m + c$
$\displaystyle c = -4 - 2m$
Hence equation of the line is $\displaystyle y = mx - 4 - 2m$
Equating this into $\displaystyle y = x^2 - 4x$,
$\displaystyle mx - 4 - 2m = x^2 - 4x$
$\displaystyle x^2 - 4x - mx + 4 + 2m = 0$
$\displaystyle x^2 - x(4 + m) + 4 + 2m = 0$
I'm not sure how I can factorise this into the answer that is $\displaystyle (2 + m, m^2 - 4)$
Any help would be greatly appreciated and thank you very much in advance! :)
• Aug 27th 2010, 11:20 PM
CaptainBlack
Quote:

Originally Posted by caramelcake
These are the steps I have done thus far:
Let $\displaystyle c$ be the $\displaystyle y$ - intercept
$\displaystyle -4 = 2m + c$
$\displaystyle c = -4 - 2m$
Hence equation of the line is $\displaystyle y = mx - 4 - 2m$
Equating this into $\displaystyle y = x^2 - 4x$,
$\displaystyle mx - 4 - 2m = x^2 - 4x$
$\displaystyle x^2 - 4x - mx + 4 + 2m = 0$
$\displaystyle x^2 - x(4 + m) + 4 + 2m = 0$
I'm not sure how I can factorise this into the answer that is $\displaystyle (2 + m, m^2 - 4)$
Any help would be greatly appreciated and thank you very much in advance! :)

Use the quadratic formula to find the roots, when you do you will find that you do not have the given solution, so check your previous work (but it looks OK to me, so may be check that you are looking at the correct solution)

CB
• Aug 28th 2010, 01:43 AM
caramelcake
Thank you CaptainBlack! But I have tried using the quadratic formula and can't seem to continue:
(I got a latex error so forgive my handwriting)
http://picpanda.com/images/zz33rm40rio1zxj1j.png
Does this mean that my answer should contain a surd?
• Aug 28th 2010, 02:09 AM
Educated
How did you get (m^2 - 16m) inside the square root? Shouldn't it be just M^2?

$\displaystyle x = \dfrac{4+m\pm\sqrt{(-4-m)^2 - 4 * 1 * (4+2m)}}{2}$

This is what I got:

$\displaystyle x = \dfrac{4+m\pm\sqrt{16 + m^2 + 8m - 16-8m}}{2}$

$\displaystyle x = \dfrac{4+m\pm\sqrt{m^2}}{2}$

$\displaystyle x = \dfrac{4+m+m}{2} = 2+m$

$\displaystyle x = \dfrac{4+m-m}{2} = 2$

Therefore the line will always hit the parabola at point 2, and also the point 2 + gradient.
• Aug 28th 2010, 03:17 AM
caramelcake
Thank you Educated! I have finished this sum now.
@CaptainBlack
Why isn't the textbook answer right then?
Taking $\displaystyle x$ to be $\displaystyle 2 + m$,
$\displaystyle y = m (2 + m) - 4 - 2m$
$\displaystyle y = 2m + m^2 - 4 - 2m$
$\displaystyle y = m^2 -4$
Hence the coordinates of point B are $\displaystyle (2 + m, m^2 -4)$
• Aug 28th 2010, 07:31 AM
CaptainBlack
Quote:

Originally Posted by caramelcake
Thank you Educated! I have finished this sum now.
@CaptainBlack
Why isn't the textbook answer right then?
Taking $\displaystyle x$ to be $\displaystyle 2 + m$,
$\displaystyle y = m (2 + m) - 4 - 2m$
$\displaystyle y = 2m + m^2 - 4 - 2m$
$\displaystyle y = m^2 -4$
Hence the coordinates of point B are $\displaystyle (2 + m, m^2 -4)$

My mistake, I misinterpreted what the text books answer was supposed to be.

CB