Question

The angle between the tangents at those points on the curve $x = {t^2} + 1{\text{ and }}y = {t^2} - t - 6$ where it meets the $x - axis$ is :
$\left( 1 \right) \pm {\tan ^{ - 1}}\left( {\dfrac{4}{{29}}} \right)$
$\left( 2 \right) \pm {\tan ^{ - 1}}\left( {\dfrac{5}{{49}}} \right)$
$\left( 3 \right) \pm {\tan ^{ - 1}}\left( {\dfrac{{10}}{{49}}} \right)$
$\left( 4 \right) \pm {\tan ^{ - 1}}\left( {\dfrac{8}{{29}}} \right)$

Answer 1

Hint: We have given the equation of curve in parametric form, since both $x{\text{ and y}}$ equation are represented by a common third variable $t$ i.e. both the variables are function of $t$ . First we will differentiate $x{\text{ and y}}$ individually and then calculate the value of $\dfrac{{dy}}{{dx}}.$ Then in the question it is given that the curve meets the $x - axis$ means the value of $y$ coordinate will be zero, we will put ${t^2} - t - 6 = 0$ and we will get two values for $t$ , using these values we will calculate the slopes for each point and then by using the angle slope formula we will calculate the angle between the tangents.

Complete step by step solution:
Given: $x = {t^2} + 1$
Differentiate the above equation w.r.t. $t$ , we get;
$ \Rightarrow \dfrac{{dx}}{{dt}} = 2t{\text{ }}......\left( 1 \right)$
Given: $y = {t^2} - t - 6$
Differentiate the above equation w.r.t. $t$ , we get;
$ \Rightarrow \dfrac{{dy}}{{dt}} = 2t - 1{\text{ }}......\left( 2 \right)$
Now, dividing the equation $\left( 2 \right)$ by equation $\left( 1 \right)$ , we get;
$ \Rightarrow \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{2t - 1}}{{2t}}$
Simplifying the above equation, we get the value of $\dfrac{{dy}}{{dx}}$ as;
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2t - 1}}{{2t}}{\text{ }}......\left( 3 \right)$
According to the question when the curve meets the $x - axis$ , then the value of $y{\text{ will be 0}}$;
$\therefore {t^2} - t - 6 = 0$
On factorization of the above equation we get;
$ \Rightarrow {t^2} - 3t + 2t - 6 = 0$
$ \Rightarrow \left( {t - 3} \right)\left( {t + 2} \right) = 0$
Therefore, the values of $t$ are $t = 3{\text{ and }}t = - 2$ .
When $t = 3$ , then by $x = {t^2} + 1$ , the value of $x = 10$ .
Similarly when $t = - 2{\text{ then }}x = 5$ .
Hence the point where the curve meets the $x - axis{\text{ are }}\left( {10,0} \right){\text{ and }}\left( {5,0} \right)$.
$\left( 1 \right)$ The slope of the tangent at point $\left( {10,0} \right)$ is given by:
$ \Rightarrow {m_1} = {\left. {\dfrac{{dy}}{{dx}}} \right]_{x = 10}} = {\left. {\dfrac{{dy}}{{dx}}} \right]_{t = 3}}$
Now, put the value of either $x{\text{ or t in equation }}\left( 3 \right)$ , we get the value of ${m_1}$ as;
$ \Rightarrow {m_1} = {\left. {\dfrac{{2t - 1}}{{2t}}} \right]_{t = 3}}$
$ \Rightarrow {m_1} = \dfrac{5}{6}{\text{ }}......\left( 4 \right)$

$\left( 2 \right)$ The slope of the tangent at point $\left( {5,0} \right)$ is given by:
$ \Rightarrow {m_1} = {\left. {\dfrac{{dy}}{{dx}}} \right]_{x = 5}} = {\left. {\dfrac{{dy}}{{dx}}} \right]_{t = - 2}}$
Now, put the value of either $x{\text{ or t in equation }}\left( 3 \right)$ , we get the value of ${m_2}$ as;
$ \Rightarrow {m_2} = {\left. {\dfrac{{2t - 1}}{{2t}}} \right]_{t = - 2}}$
$ \Rightarrow {m_2} = \dfrac{5}{4}{\text{ }}......\left( 5 \right)$
We know that when two lines intersect each other at a point then the angle between them can be expressed in terms of their slopes and the angle is given by the formula;
$ \Rightarrow \tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
Where ${m_1}{\text{ and }}{{\text{m}}_2}$ are slopes of the tangents respectively and $\theta $ is the angle between the two tangents;
Now put the value of ${m_1}{\text{ and }}{m_2}{\text{ from equation }}\left( 4 \right){\text{ and equation }}\left( 5 \right),$ in the angle formula we get;
$ \Rightarrow \tan \theta = \left| {\dfrac{{\dfrac{5}{4} - \dfrac{5}{6}}}{{1 + \dfrac{{25}}{{24}}}}} \right|$
$ \Rightarrow \tan \theta = \left| {\dfrac{{\dfrac{{15 - 10}}{{12}}}}{{\dfrac{{24 + 25}}{{24}}}}} \right|$
Further simplifying the above equation, we get;
$ \Rightarrow \tan \theta = \left| {\dfrac{{10}}{{49}}} \right|$
Therefore, the value of $\theta $ can be given by;
$ \Rightarrow \theta = \pm {\tan ^{ - 1}}\left( {\dfrac{{10}}{{49}}} \right)$
Hence the value of the angle between the tangents is $\theta = \pm {\tan ^{ - 1}}\left( {\dfrac{{10}}{{49}}} \right)$ .
Therefore the correct answer for this question is option $\left( 3 \right)$ .

So, the correct answer is “Option 3”.

Note: The equation of curve in parametric form is represented to make the calculations easier (they are easier to differentiate) in case of complex equations of the curves . There are some noticeable points about the angle slope formula : $\left( 1 \right)If\tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$ is positive then the angle between the tangent lines is acute. $\left( 2 \right)If\tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$ is negative then the angle between the lines is obtuse.