A curve is given by the equations $x=at^2,y=at^3 $.A variable pair of perpendicular lines through the origin 'O' meet the curve at $P$ & $Q$.Show that the locus of the point of intersection of the tangents at $P$ & $Q$ is $4y^2=3ax-a^2$
What I have done:I have found the general equation of pair of perpencicular lines passing through origin with slope $m$.The general equation comes out to be $\to$ $x^2+mxy=y^2+\frac{xy}{m}$,Now I have found the equation of tangent at any two general points on the given curve.The equations are
$(1)\to y-at_1^3=\frac{3t_1}{2}(x-at_1^2)$ $(2)\to y-at_2^3=\frac{3t_2}{2}(x-at_2^2)$
Now,I don't know what to do further and question looks very messy with 3 variables coming in .Please help
It wont be this messy if instead of parametric form, you solve it by finding the curve, $$ay^2=x^3$$,take two lines passing through the origin as $y=mx$ and $y=\frac{-x}{m}$, find the points P and Q, $$P (am^2,am^3) , Q(\frac{a}{m^2},\frac{-a}{m^3})$$ Next find the slope of the tangent at these points,at P, slope is $\frac{3m}{2}$, at Q it is $\frac{-3}{2m}$, The equations of the tangents are, $$2y=3mx-am^3$$ and $$2y=\frac{-3x}{m}+\frac{a}{m^3}$$ Solving these, I get $$x=\frac{a}{3}[(m-\frac{1}{m})^2+1]$$ and $$y=-\frac{a}{2}(m-\frac{1}{m})$$ By the second equation,$ (m-\frac{1}{m})=\frac{-2y}{a}$
Substituting this in the first, gives $$4y^2=3ax-a^2$$ the desired result.