Locus of intersection of perpendicular tangents to the curve 4y3 = 27x2
6212y2dy/dx=54x
hence,
\frac{dy}{dx}=\frac{9x_1}{2y_1^2}\\
Equation of the line is given by :
y-y_1 = (x-x_1)\frac{9x_1}{2y_1^2}
and
y-y_2 = (x-x_2)\frac{9x_2}{2y_2^2}
multiplying slopes at 2 points x1,y1 and x2, y2 such that they are perpendicular
\frac{9x_1}{2y_1^2}\frac{9x_2}{2y_2^2}=-1\\
Is it becoming to dirty...
seems to me atleast :(
1we get dy/dx= 9x/2y2
thus equation of tangent = y-k/x-h=9h/2k2
we know m=9h/2k2
and further 4k3=27h2
so we can express them in terms of m
the final equation we get= m3x - m2y + 1=0
let y, x be the intersection of perpendicular tangents
then this equation will have two roots m1 and m2
such that m1m2=-1
let the other root be m3
m1m2m3=-1/x
m1m2 + m2m3 +m3m1 = 0
m1 +m2 +m3 = y/x
solving we get locus as y=x2 + 1