# Find the equation of the tangent to the curve $ay^{2}= x^{3}$ at the point $\left ( am^{2} ,am^{3}\right )\cdot$

curve $ay^{2}= x^{3}$
point $\left ( am^{2},am^{3} \right )$
we have a family of curves defined for a>0 by
$C\left ( a \right )= ay^{2}= x^{3}---(A)$
or in parametric form
$C\left ( a \right )\left\{\begin{matrix} x\left ( m \right )= am^{2} & \\ y\left ( m \right ) = am^{3}& \end{matrix}\right.$

Here we have a plot for a =1
If we differentiate equation (A) implicitly w.r.t x we have:
$2ay\frac{dy}{dx}= 3x^{2}\Rightarrow \frac{dy}{dx}= \frac{3x^{2}}{2ay}$
The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. so, at a generic point P(am2,am3) the derivative is
$m_{T}= \frac{3x^{2}}{2ay}= \frac{3\left ( am^{2} \right )^{2}}{2a\left ( am^{3} \right )}= \frac{3a^{2}m}{2a^{2}m^{3}}= \frac{3}{2}m$
So the tangent point through P(am2 ,am3) and has gradient $m_{T}= \frac{3}{2}m,$ using the point/slope form $y-y_{1} = m\left ( x-x_{1} \right )$ the tangent equation is
$y-am^{3}= \frac{3}{2}m\left ( x-am^{2} \right )$
$y-am^{3}= \frac{3}{2}mx-\frac{3}{2}am^{3}$
$\therefore y= \frac{3}{2}mx\, \: \frac{-1}{2}am^{3}$

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