# Locus of point of intersection of the tangents which are at right angles

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.

$y=m_1(x+1)+1/m_1$ and $y=m_2(x+2)+2/m_2$

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 $x=-3$.

HINT:

The parametric form of $y^2=4(x+1)\ \ \ \ (1),y^2=8(x+2)\ \ \ \ (2)$ are $(t^2-1,2t);(2u^2-2,4u)$

So, the equation of the tangent of $(1)$ at $y=2t$ will be $$y(2t)=4\cdot\dfrac{(x+t^2-1)}2+4\iff t^2-t(y)+x+1=0\ \ \ \ (3)$$

and that of $(2)$ at $y=4u$ will be $$x-yu+2+2u^2=0\ \ \ \ (4)$$

So. we need $$\dfrac1{tu}=-1\iff u=-\dfrac1t$$

Consequent, $(4)$ becomes $$t^2(x+2)+t(y)+2=0\ \ \ \ (5)$$

Solve $(3),(5)$ for $t,t^2$

• Since they are perpendicular,the best method would be to use the equation of director circle for parabola,this method is really complicated and not suited for competitive examinations like IIT – GK A Nov 17 '18 at 16:26