# finding tangent

• May 15th 2009, 02:42 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, 03:01 PM
Scott H
Hint: If

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

then

\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, 03:02 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 $x = \alpha$ into the given equation to get the y-coordinate. So you have a point on the tangent.

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

Substitute all of the above into $y - y_1 = m (x - x_1)$ to get the equation.
• May 15th 2009, 07:52 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, 03:46 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.