Question

Angle between the tangents drawn to $y^2=4x$ at the points where it is intersected by the line $y=x-1$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

$y^2=4x$ and $y=x-1$

$y^2=(x-1{)}^2=4x$

$x^2-2x+1=4x$

$x^2-6x+1=0$

$\frac{-(-6)\pm \sqrt{36-4\left(1\right)\left(1\right)}}{2\left(1\right)}$

$\frac{6\pm 32}{2}=3\pm 2\sqrt{2}$

$x_1=3+2\sqrt{2}$, $x_2=3-2\sqrt{2}$

$y_1=2+2\sqrt{2}$, $y_2=2-2\sqrt{2}$

$2y\frac{dy}{dx}=4$

$m_1=\frac{2}{y_1},m_2=\frac{2}{y_2}$

$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

$=\frac{{\frac{2}{y}}_1-{\frac{2}{y}}_2}{1+\left({\frac{2}{y}}_1\right)\left({\frac{2}{y}}_2\right)}$

$=\frac{2y_2-2y_1}{y_1{y}_2+4}$

$=\frac{2[2-2\sqrt{2}-2-2\sqrt{2}]}{4-8+4}$

$\tan \theta =\infty$

$\theta ={90}^o=\frac{\pi }{2}$.