# finding tangent

• May 15th 2009, 01:42 PM
slaypullingcat
finding tangent
Show that the tangent to P: y= a(x-h)^2 + k at point T, where x= alpha, is y=2a(alpha-h)x+k-a(alhpa^2-h^2).

Thank you for any help
• May 15th 2009, 02:01 PM
Scott H
Hint: If

$\displaystyle y=a(x-h)^2+k,$

then

\displaystyle \begin{aligned} y'&=\frac{d}{dx}(a(x-h)^2+k)\\ &=\frac{d}{dx}(a(x-h)^2)+\frac{d}{dx}(k)\\ &=2a(x-h)\cdot \frac{d}{dx}(x-h)+0\;\;\;\;\;\;\;\;\;\;\mbox{(Chain Rule)}\\ &=2a(x-h)\cdot 1\\ &=2a(x-h). \end{aligned}
• May 15th 2009, 02:02 PM
mr fantastic
Quote:

Originally Posted by slaypullingcat
Show that the tangent to P: y= a(x-h)^2 + k at point T, where x= alpha, is y=2a(alpha-h)x+k-a(alhpa^2-h^2).

Thank you for any help

Substitute $\displaystyle x = \alpha$ into the given equation to get the y-coordinate. So you have a point on the tangent.

Calculate the value of dy/dx at $\displaystyle x = \alpha$ to get the gradient of the tangent.

Substitute all of the above into $\displaystyle y - y_1 = m (x - x_1)$ to get the equation.
• May 15th 2009, 06:52 PM
slaypullingcat
I am using m=2a(x-h), x1=alpha and y1=a(alpha-h)^2 +k. I'm getting fairly close but its still not correct. Are my substitution values right? If so I guess I am simplifying incorrectly
• May 16th 2009, 02:46 AM
mr fantastic
Quote:

Originally Posted by slaypullingcat
I am using m=2a(x-h), x1=alpha and y1=a(alpha-h)^2 +k. I'm getting fairly close but its still not correct. Are my substitution values right? If so I guess I am simplifying incorrectly

Your trouble will be somewhere in the algebra.