Question

The angle of intersection of intersection of the curves $y = {x^2}$ and $x = {y^2}$ at $(1,1)$ is
A). ${\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right)$
B). ${\tan ^{ - 1}}(1)$
C). $90^\circ$
D). ${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$

Answer 1

Here we are asked to find the angle of intersection of two given curves $y = {x^2}$ and $x = {y^2}$ at a point $(1,1)$ . First, we need to find the slopes of the given two curves. Then we will use the formula of the angle of intersection of two curves and substituting the slopes of the curves to find the angle between them.
Formula Used: The formula that we need to know before solving the problem:
If ${m_1}$ and ${m_2}$ are the slopes of the two curves $y = {f_1}(x)$ and $y = {f_2}(x)$ then the angle of intersection of these curves is $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_{}}}}$.

Complete step-by-step solution:
We aim to find the angle of intersection of the curves $y = {x^2}$ and $x = {y^2}$ at a point $(1,1)$.
First, we have to find the slopes of the given two curves.
Let us consider the first given curve $y = {x^2}$. Let ${m_1}$ be the slope of this curve.
On differentiating the above curve, we get
$\dfrac{{dy}}{{dx}} = 2x$
Thus, we have found the slope of the first curve. Now we will find the slope at a given point $(1,1)$.
${\left( {\dfrac{{dy}}{{dx}}} \right)_{(1,1)}} = 2(1) = 2$
Therefore, ${m_1} = 2$
Now let us find the slope of the other curve $x = {y^2}$. Let ${m_2}$ be the slope of this curve.
On differentiating the above curve, we get
$1 = 2y\dfrac{{dy}}{{dx}}$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{2y}}$
Thus, we have found the slope of first curve. Now we will find the slope at a given point $(1,1)$.
${\left( {\dfrac{{dy}}{{dx}}} \right)_{(1,1)}} = \dfrac{1}{{2(1)}} = \dfrac{1}{2}$
Therefore, ${m_2} = \dfrac{1}{2}$
Now we have found the slopes of both given curves. Now let us find the angle of intersection of these curves.
We know that if ${m_1}$ and ${m_2}$ are the slopes of the two curves $y = {f_1}(x)$ and $y = {f_2}(x)$ then the angle of intersection of these curves is $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}$.
Here the ${m_1} = 2$ and ${m_2} = \dfrac{1}{2}$ are the slopes of the two curves $y = {x^2}$ and $x = {y^2}$ then the angle of intersection of these curves at a point $(1,1)$ is
$\tan \theta = \dfrac{{2 - \dfrac{1}{2}}}{{1 + (2)\left( {\dfrac{1}{2}} \right)}}$
On simplifying this we get
$\tan \theta = \dfrac{{\dfrac{3}{2}}}{{1 + 1}}$
$\Rightarrow \tan \theta = \dfrac{3}{4}$
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
Thus, the angle of intersection of the given two curves is $\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$.
Now let us see the options, option (a) ${\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right)$ is an incorrect option since we got the angle as $\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
Option (b) ${\tan ^{ - 1}}(1)$ is an incorrect option since we got the angle as $\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
Option (c) $90^\circ$ is an incorrect option since we got the angle as $\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
Option (d) ${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$ is the correct option as we got the same angle in our calculation above.
Hence, option (d) ${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$ is the correct answer.

Note: In the given options, option (c) is given as $\theta = 90^\circ$ if we find the tangent value of this degree, we get $\tan 90^\circ = \infty$ which is undefined but from our calculation, we have to get $\dfrac{3}{4}$ . Thus, option (c) cannot be the right answer.