Question

For a parabola, if $L_1:x=y+1$ is the axis of symmetry, $L_2:x+y=5$ is tangent at vertex and $L_3:y=4$ is a tangent at a point $P,$ then the equation of circumcircle of the triangle formed by the tangent and normal at point $P$ and axis of parabola is

• A. $x^2+y^2+2y=31$
• B. $x^2+y^2-2y=31$
• C. $x^2+y^2+2x=31$
• D. $x^2+y^2-2x=31$

Answer 1

The correct option is D $x^2+y^2-2x=31$

$L_1:x=y+1$ is axis of symmetry.
$L_2:x+y=5$ is tangent at vertex.
$L_3:y=4$ is tangent at $P$
Vertex is given by point of intersection of $L_1$ and $L_2$
$2y+1=5\Rightarrow y=2\Rightarrow x=3\text{Vertex}=V=(3,2)$

As focus lies on $L_1,$ let it be $(h,h-1)$
Foot of perpendicular from focus on $L_3$ lies on the tangent at vertex $L_2$, so the point of intersection of $L_2$ and $L_3$ is $(1,4)$
Slope of line joining $(h,h-1)$ and $(1,4)$ tends to infinity,
$m=\frac{h-5}{h-1}\therefore h=1\text{Focus}=S=(1,0)$
We know that $V$ is midpoint of $S$ and $A$, so
$A=(5,4)$
Directrix is parallel to $L_2$ and passing through $A$, so its equation is
$x+y=9$
Let point $P=(k,4)$
Using the definition of parabola,
$SP=PM\Rightarrow \sqrt{(k-1{)}^2+16}=\left|\frac{k+4-9}{\sqrt{2}}\right|\Rightarrow 2(k^2-2k+17)=(k^2-10k+25)\Rightarrow k^2+6k+9=0\Rightarrow k=-3\therefore P=(-3,4)$
Normal at $P$
$x=-3$


$B=(-3,-4)$
Equation of circumcircle of triangle $PAB$, where $AB$ is diameter, is
$(x-5)(x+3)+(y-4)(y+4)=0\Rightarrow x^2+y^2-2x=31$