Application of derivative - finding the point on curve Problem : 

Find the point on the curve $ay^2=x^3$ in the fourth quadrant where normal to the curve makes equal non zero intercepts with axes. 

Solution : 

Let the point at which normal is drawn can be given by $(x_1,y_1)$ Then it must satisfy $ay^2=x^3$

i.e. $ay_1^2=x_1^3$ 

= $y_1 =\pm \sqrt{\frac{x_1^3}{a}}$

Now differentiating the  given equation we get : 

$\frac{dy}{dx} =\frac{3x^3}{2ay}$ therefore $\frac{dy}{dx}$ at $(x_1,y_1) = \pm \frac{3}{2}\sqrt{\frac{x_1}{a}}$

Please guide further to clear the concepts.... thanks a lot..
HINT: The $\frac{dy}{dx}$ gives you slope of the tangent at that point [Differentiate correctly]. So find the slope of the normal at that point. Now you can write down the equation of the normal(point-slope form).
Further apply the condition on equal non zero intercepts.
Slope of tangent is $+1$ or $-1$,in $4^{\text{th}}$ quad, $x&gt;0$, $y&lt;0$ therefore find $dy/dx$ and put it equal to $\pm1$,  you get $x=4a/3$ and $y=-a(4/3)^{3/2}
$.