Question

Find the angle between tangent of the curve $y=(x+1)(x-3)$ at the point where it cuts the axis of $x$.

• A. ${\tan }^{-1}\left(\frac{8}{15}\right)$
• B. ${\tan }^{-1}\left(\frac{15}{8}\right)$
• C. ${\tan }^{-1}4$
• D. $Noneofthese$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{8}{15}\right)$

The curve $y=(x+1)(x-3)=x^2-2x-3$.........(1) cuts the axis of $x$ at $(-1,0)$ and $\left(3.0\right)$. These points are obtained by putting $y=0$ in the equation (1).

Now the slope of the tangent to the curve (1) at $(-1,0)$ be $\left(m_1\right)={\frac{dy}{dx}|}_{(-1,0)}=[2x-2{]}_{(-1,0)}=-4$.

Again the slope of the tangent to the curve (1) at $(-1,0)$ be $\left(m_2\right)={\frac{dy}{dx}|}_{(3,0)}=[2x-2{]}_{(3,0)}=4$.

Now the angle between the tangents to (1) at those points be

${\tan }^{-1}\left(\frac{m_1-m_2}{1+m_1.m_2}\right)$

$={\tan }^{-1}\left(\frac{8}{15}\right)$.

So the required angle be ${\tan }^{-1}\left(\frac{8}{15}\right)$.