Stated question: The point T ($at^2$, $2at$) lies on the parabola $y^2=4ax$ and L is the point ($-a$, $2a$). M is the mid-point of TL. Find the equation of the locus of M as T moves on the parabola.
Given solution: M will have the coordinates $(\frac{at^2-a}{2}, \frac{2at+2a}{2}) = (\frac{a}{2}(t^2-1), a(t+1))$. As T moves on the parabola the parameter varies and the parametric equation of the locus of M will be: $x=\frac{a}{2}(t^2-1)$, $y=a(t+1)$.
Eliminating $t$ from these equations gives $y^2=4(x+y)$, the cartesian equation of the locus of M.
My problem: When I try to eliminate $t$, I get: $x=\frac{a}{2}((\frac{y-a}{a})^2-1)=\frac{ay^2-2a^2y+a^3}{2a^2}-\frac{a}{2}=\frac{y^2-2ay}{2a}\therefore y^2=2a(x+y)$.
How am I supposed to eliminate $a$ and reach the given solution, $y^2=4(x+y)$?
Best Answer
The solution you found is the right one. It will have to depend on $a$. Are you sure you looked at all the assumptions? If you did, then the given solution isn't right, and your solution is right.