Question

let $P$ be a variable point. From $P,PQandPR$ are tangents drawn to the parabola $y^2=4x$. If $\angle QPR$ is always $45$, then locus of $P$ is

• A. $x^2-y^2+6x+1=0$
• B. $x^2{+y}^2+6x+1=0$
• C. $x^2{-y}^2-6x+1=0$
• D. $Noneofthese$

Answer 1

The correct option is A $x^2-y^2+6x+1=0$

Let $(t_1^2,2t_1)$ be the co-ordinates of point Q and $(t_2^2,2t_2)$ be the co-ordinates of point R

Equation of tangent at Q is $t_{1}y=x+t_1^2$

Equation of tangent at R is $t_{2}y=x+t_2^2$

These two lines intersect at the point $(t_1{t}_2,t_1+t_2)$.

So the co-ordinates of point P is $(h,k)=(t_1{t}_2,t_1+t_2)$.

Slope of line PQ $=m_1=\frac{1}{t_1}$

Slope of line PR $=m_2=\frac{1}{t_2}$

Angle between these lines is ${45}^0$

$\therefore tan\left(45\right)=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\therefore 1=\left|\frac{\frac{1}{t_1}-\frac{1}{t_2}}{1+\frac{1}{t_1}\frac{1}{t_2}}\right|$

$\therefore (t_2-t_1{)}^2=(t_1{t}_2+1{)}^2$

$\therefore (t_2+t_2{)}^2-4t_1{t}_2=(t_1{t}_2+1{)}^2$

$\therefore k^2-4h=(h+1{)}^2$

$\therefore h^2-k^2+6h+1=0$

So, the locus of point P is $x^2-y^2+6x+1=0$

So, the answer is option A