Question

The angle between the tangents at those points on the curve $x=t^2+1$ and $y=t^2-t-6$, where it meets $X-$ axis is

• A. $\pm {\tan }^{-1}\left(\frac{4}{29}\right)$
• B. $\pm {\tan }^{-1}\left(\frac{5}{49}\right)$
• C. $\pm {\tan }^{-1}\left(\frac{10}{49}\right)$
• D. $\pm {\tan }^{-1}\left(\frac{8}{29}\right)$

Answer 1

The correct option is C

$\pm {\tan }^{-1}\left(\frac{10}{49}\right)$

Explanation for the correct option:

Equation of the given curve in parametric form,
$x=t^2+1$ and $y=t^2-t-6$
$Y-$ coordinate of the point, where the given curve meets $X-$ axis is zero.
When $$\begin{gathered} y=0 \\ \Rightarrow t^2-t-6=0 \end{gathered}$$

on simplifying, we get
$$\begin{gathered} \Rightarrow t^2-3t+2t-6=0 \\ \Rightarrow t(t-3)+2(t-3)=0 \\ \Rightarrow (t-3)(t+2)=0 \\ \Rightarrow t=3,-2 \end{gathered}$$

Hence, the points where the curve meets the $X-$-axis are $(10,0),(5,0).$
Now, $\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{2t-1}{2t}$
The slope of the tangent at point $(10,0)$
$m_1={\left[\frac{dy}{dx}\right]}_{x=10}={\left[\frac{dy}{dx}\right]}_{t=3}=\frac{5}{6}$
The slope of the tangent at point (5, 0),
​$m_2={\left[\frac{dy}{dx}\right]}_{x=5}={\left[\frac{dy}{dx}\right]}_{t=-2}=\frac{5}{4}$
If $\theta$ be the angle between two tangents, then
$$\begin{gathered} \tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right| \\ =\left|\frac{{\displaystyle \frac{5}{4}}-{\displaystyle \frac{5}{6}}}{1+\frac{5}{4}.\frac{5}{6}}\right| \\ =\left|\frac{10}{49}\right| \\ \tan \theta =\pm \frac{10}{49} \\ \Rightarrow \theta =\pm {\tan }^{-1}\left(\frac{10}{49}\right) \end{gathered}$$

Hence, the correct option is (C)