Question

Let $B(16,-16)$ be an end point of a focal chord $AB$ of the parabola $y^2=16x$ and let the tangent and normal at $A$ meet the axes in $T$ and $N$ respectively. Then reflection of the circle through $A,T$ and $N$ about the line $y=x$, is

• A. $x^2+y^2-8x+9=0$
• B. $x^2+y^2-8y-9=0$
• C. $x^2+y^2+8x+9=0$
• D. $x^2+y^2+8x-9=0$

Answer 1

The correct option is B $x^2+y^2-8y-9=0$

Let the end points of focal chord be $A(a{t}_1^2,2a{t}_1)$ and $B(a{t}_2^2,2a{t}_2)$
$\because$ $AB$ is a focal chord, $t_1{t}_2=-1\Rightarrow A(1,4)$
Now focus of the parabola is $S(4,0)$.
Tangent at point $A(1,4)$ is $2x-y+2=0$ which meets x axis $T(-1,0)$
Normal at point $A(1,4)$ is $x+2y-9=0$ which meets x axis at $N(9,0)$
Now of circle passes through points $A(1,4),T(-1,0)$ and $N(9,0)$
Let the center of this circle be $C(p,q)$, distance of points $A,T,N$ are same and will be equal to radius.
$AC=TC=NC$
$(p-1{)}^2+(q-4{)}^2=(p+1{)}^2+q^2=(p-9{)}^2+q^2$
$\Rightarrow p=4,q=0$
$\therefore$ center $=C(4,0)$, which is the focus $A$
Radius of circle $=AC=\sqrt{(p-1{)}^2+(q-4{)}^2}=5$
Equation of circle with Focus as the center and radius $=5$ is
$(x-4{)}^2+(y-0{)}^2=5^2$
Now reflection of this circle about the line $y=x$ is
$(x-0{)}^2+(y-4{)}^2=5^2$
$\Rightarrow x^2+y^2-8y-9=0$