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
Hint: If
$\displaystyle y=a(x-h)^2+k,$
then
$\displaystyle \begin{aligned}
y'&=\frac{d}{dx}(a(x-h)^2+k)\\
&=\frac{d}{dx}(a(x-h)^2)+\frac{d}{dx}(k)\\
&=2a(x-h)\cdot \frac{d}{dx}(x-h)+0\;\;\;\;\;\;\;\;\;\;\mbox{(Chain Rule)}\\
&=2a(x-h)\cdot 1\\
&=2a(x-h).
\end{aligned}$
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.