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

Answer: (C)

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 0.
When y = 0, then $t^2-t-6=0$
$\Rightarrow t^2-3t+2t-6=0$
$\Rightarrow t(t-3)+2(t-3)=0$
$\Rightarrow (t-3)(t+2)=0$
$\Rightarrow$ t = 3 or - 2
when t = 3, then x = 10
when t = -2, then x = 5
Hence, the points where the curve meets the X-axis are (10, 0) and (5, 0).
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t-1}{2t}$
Slope of the tangent at point (10, 0)
$m_1=\frac{dy}{dx}{]}_{x=10}=\frac{dy}{dx}{]}_{t=3}=\frac{5}{6}$
Slope of the tangent at point (5, 0),
$m_2=\frac{dy}{dx}{]}_{x=5}=\frac{dy}{dx}{]}_{t=-2}=\frac{-5}{-4}=\frac{5}{4}$
If $\theta$ be the angle beween two tangents, then
$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|=\left|\frac{\frac{5}{4}-\frac{5}{6}}{1+\frac{5}{6}.\frac{5}{4}}\right|$
$=\left|\frac{\frac{15-10}{12}}{\frac{24+25}{24}}\right|=\left|\frac{\frac{5}{12}}{\frac{49}{24}}\right|=\left|\frac{10}{49}\right|$
$\Rightarrow \tan \theta =\pm \frac{10}{49}$
$\therefore \theta ={\tan }^{-1}\left(\pm \frac{10}{49}\right)=\pm {\tan }^{-1}\left(\frac{10}{49}\right)$