Calculating tangent at two points $(4a,8a)$

Question

Find the equation of the tangent to the curve $ay^2=x^3$ at the points $(4a,8a)$

I have re-arranged the equation to get $$y = \left(\frac{x^3}{a}\right)^{\frac{1}{2}}$$

Then taking its derivative I get $$\frac{dy}{dx}=\frac{3\left(\sqrt{\frac{x^3}{a}}\right)}{2x}$$, when $x=4a$, I find that $\frac{dy}{dx}=3$

and so I have tried: $$\frac{y-\left(\frac{x^3}{a}\right)^{\frac{1}{2}}}{x-4a}=3$$

However, this will not give me a simple answer which is to be $y = 3x-4a$

How do I correctly calculate at both points?

Answer 1

You have probably miscalculated the derivative when the chain rule was applied.

Note that $\dfrac{du^\frac{1}{2}}{du}=\dfrac{1}{2}u^{-\frac{1}{2}}$.

Hence,

$y'=\dfrac{d}{d\dfrac{x^3}{a}}(\dfrac{x^3}{a})^\frac{1}{2}\cdot\dfrac{d}{dx}(\dfrac{x^3}{a})$

$=\dfrac{1}{2}\cdot(\dfrac{x^3}{a})^{-\frac{1}{2}}\cdot\dfrac{3x^2}{a}$

$=\dfrac{3}{2}\sqrt{\dfrac{x}{a}}$

$y'|_{x=4a}=3$

Therefore, the equation of tangent:

$\dfrac{y-8a}{x-4a}=3$

$\underline{\underline{y=3x-4a}}$

Answer 2

You need to use the implicit derivation method: in your case is $$2ayy’=3x^2 \implies y’=\frac{3x^2}{2ay}$$ and then proceed as usual.