Find the locus of the point of intersection of the tangents to the parabolas $y^2=4(x+1)$ and $y^2=8(x+2)$ which are at right angles.
Equation of tangents will be
$y=m_1(x+1)+1/m_1$ and $y=m_2(x+2)+2/m_2$
Now even if I use the condition that $m_1 \cdot m_2=-1$, I am not able to eliminate $m_1$. Please help me with this.
Best Answer
You already have
Now the $x$-coordinate $X$ of the intersection point of the tangent lines is given by $$X=\frac{2m_1m_2^2+2m_1-m_1^2m_2-m_2}{m_1m_2(m_1-m_2)}$$ Now using $m_1m_2=-1$ gives $$X=\frac{-2m_2+2m_1+m_1-m_2}{-(m_1-m_2)}=\frac{3(m_1-m_2)}{-(m_1-m_2)}=-3$$ and so the $y$-coordinate $Y$ of the intersection point is given by $$Y=-2m_1+\frac{1}{m_1}$$ with $$\lim_{m_1\to +\infty}Y=-\infty,\quad \lim_{m_1\to 0^+}Y=+\infty$$ Therefore, the locus we want is the line $x=-3$.