Question
Tangent and normal are drawn at $$P\left( {16,\,16} \right)$$ on the parabola $${y^2} = 16x,$$ which intersect the axis of the parabola at $$A$$ and $$B,$$ respectively. If $$C$$ is the centre of the circle through the points $$P,\,A$$ and $$B$$ and $$\angle CPB = \theta ,$$ then a value of $$\tan \,\theta $$ is :
A.
$$2$$
B.
$$3$$
C.
$$\frac{4}{3}$$
D.
$$\frac{1}{2}$$
Answer :
$$2$$
Solution :
Equation of tangent at $$P\left( {16,\,16} \right)$$ is given as :
$$x - 2y + 16 = 0$$
Slope of $$PC\left( {{m_1}} \right) = \frac{4}{3}$$
Slope of $$PB\left( {{m_2}} \right) = - 2$$
Hence, $$\tan \,\theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}.{m_2}}}} \right| = \left| {\frac{{\frac{4}{3} + 2}}{{1 - \frac{4}{3}.2}}} \right|$$
$$ \Rightarrow \tan \,\theta = 2$$
Equation of tangent at $$P\left( {16,\,16} \right)$$ is given as :
$$x - 2y + 16 = 0$$
Slope of $$PC\left( {{m_1}} \right) = \frac{4}{3}$$
Slope of $$PB\left( {{m_2}} \right) = - 2$$
Hence, $$\tan \,\theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}.{m_2}}}} \right| = \left| {\frac{{\frac{4}{3} + 2}}{{1 - \frac{4}{3}.2}}} \right|$$
$$ \Rightarrow \tan \,\theta = 2$$