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

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- May 15th 2009, 01:42 PMslaypullingcatfinding 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 PMScott 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 PMmr fantastic
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 PMslaypullingcat
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 AMmr fantastic