Question

If tangents are drawn at the points $A(1,2),B(4,-4)$ and a variable point ${}^{\prime }{C}^{\prime }$ lies on the parabola $y^2=4x$, then the locus of the orthocentre of triangle formed by these tangents is

• A. $x=0$
• B. $x-1=0$
• C. $x+1=0$
• D. $y=0$

Answer 1

The correct option is C $x+1=0$

For the given parabola $a=1$
Now,
$A\equiv (1,2)\equiv (t_1^2,2t_1)\Rightarrow t_1=1B\equiv (4,-4)\equiv (t_2^2,2t_2)\Rightarrow t_2=-2$
Let $C\equiv (t^2,2t)$

Intersection point of tangents at $A$ and $B$
$=(t_1{t}_2,(t_1+t_2\left)\right)\equiv (-2,-1)$
Let $P\equiv (-2,-1)$

Intersection point of tangents at $C$ and $B$
$=(-2t,t-2)$
Let $R\equiv (-2t,t-2)$

Intersection point of tangents at $C$ and $A$
$=(t,t+1)$
Let $Q\equiv (t,t+1)$


Slope of $QR=\frac{1}{t}$
Slope of $PQ=1$

Hence, the equation of altitiude through $R$ on $PQ$ is
$y+2-t=-(x+2t)\Rightarrow x+y+t+2=0\cdots \left(1\right)$
The equation of altitiude through $P$ on $QR$ is
$y+1=-t(x+2)\cdots \left(2\right)$

Therefore, the orthocentre is the point of intersection of equation $\left(1\right)$ and $\left(2\right)$
$-x-t-2=-tx-2t-1\Rightarrow x(t-1)+t-1=0\Rightarrow (x+1)(t-1)=0\Rightarrow x+1=0(\because t\ne 1)$

$\therefore$ The locus of orthocentre is
$x+1=0$


Alternate Solution:
For any triangle formed by the tangents drawn at $t_1,t_2,t_3$ on the parabola $y^2=4ax$ it's orthocentre will always lies on directrix, i.e. $x+a=0$
Hence, the locus is $x+1=0$