There was a Q in my class,
"From any point on $ b^4x+2a^2y^2=0$ pair of tangents PQ and PR are drawn to hyperbola $ x^2/a^2-y^2/b^2=1$. Prove that QR touches a fixed parabola and also find it's equation. (a and b are constant)"
Our teacher explained as-> let the point be (h,k). It lies on first curve hence $b^4h+2a^2k^2=0$. Writing equation of chord of contact from that point to hyperbola, $ hx/a^2-yk/b^2=1$ , now with these 2 equations eliminating h, we're left with $2xk^2+b^2yk+b^4=0$ .
I understood all that but the next part I don't understand. He said the equation of parabola can be obtained by treating the last relation as quadratic in k and equating it's discriminant with zero. Hence, he said, $ b^4y^2=8b^4x, => y^2=8x$ is the equation of parabola asked.
Any help on why this gives correct answer will be great.
That parabola is the envelope of a family of lines, parametrised by $k$. To find the equation of the envelope one must, in general, eliminate the parameter between the given equation $f(x,y,k)=0$ and its derivative $\partial f(x,y,k)/\partial k=0 $ with respect to the parameter.
On the other hand, if a polynomial shares a root with its derivative, then it is (at least) a double root. In the very special case of a quadratic polynomial, having a double root implies a vanishing discriminant: that explains why the method of your teacher works.