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

 

 

 

 
 
 
 
 

Answers (1)

           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}
 

Posted by

Ravindra Pindel

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