Question

Through the vertex $O$ of the parabola $y^2=4ax$, two chords $OP$ and $OQ$ are drawn and the circles on $OP$ and $OQ$ as diameters intersect in $R.$ If ${\theta }_1,{\theta }_2$ and $\varphi$ are the angles made with axis by the tangents at $P$ and $Q$ on the parabolas and by $OR,$ then the value of, $\cot {\theta }_1+\cot {\theta }_2$

• A. $-2\tan \varphi$
• B. $-2\tan (\pi -\varphi )$
• C. $0$
• D. $2\cot \varphi$

Answer 1

The correct option is A $-2\tan \varphi$

$ty=x+a{t}^2$

Let $t_1$ be the parameter of point P and $t_2$ be the Parameter of point Q then,

Equation of tangent at P:-

$t_{1}y=x+a{t}_1^2$

$y=\frac{1}{t_1}x+a{t}_1$

slope $m_1=\frac{1}{t_1}=tan{\theta }_1$

Similarly,

Slope $m_2=\frac{1}{t_2}=tan{\theta }_2$

$cot{\theta }_1+cot{\theta }_2=t_1+t_2$

Equation of circle at P:-

$x(x-a{t}_1^2)+y(y-2a{t}_1)=0$

$x^2+y^2-a{t}_1^{2}x-2a{t}_{1}y=0$

Equation of circle at Q:-

$x^2+y^2-a{t}_2^{1}x-2a{t}_{2}y=0$

Equation of common chord:-

$2(-a{t}_1^2+a{t}_2^2)x+2(-2a{t}_1+2a{t}_2)y=0$

$(t_2^2-t_1^2)x+2(t_2-t_1)y=0$

$y=\frac{-(t_2+t_1)x}{2}$

$tan\varphi =\frac{-(t_2+t_1)}{2}$

so, $tan\varphi =\frac{-(cot{\theta }_1+cot{\theta }_2)}{2}$

$\Rightarrow cot{\theta }_1+cot{\theta }_2=-2tan\varphi$