Derivative as a Rate Measure, Tangents and Normals
6.0 Angle of intersection of two curves
6.0 Angle of intersection of two curves
Till now we have studied the concept of tangents and normals. Now, we apply those concepts in finding the angle of intersection of two curves. As we know when two curves intersect with each other, we get atleast one point of intersection.
To find the angle of intersection of two curves, we draw tangents to both the curves at that point of intersection.
Angle of intersection of two curves is defined as the acute angle between the tangents to the two curves at their point of intersection.
Let us assume, ${C_1}$ and ${C_2}$ be two curves with equations $y=f(x)$ and $y=g(x)$ respectively. Both curves intersect at point $P(\alpha ,\beta )$ and the angle between tangents be $\theta $.
We draw two tangents at point $P$ to both the curves ${C_1}$ and ${C_2}$ say ${PT_1}$ and ${PT_2}$ respectively.
${m_1}$ be the slope of tangent ${PT_1}$ which makes an angle ${\theta _1}$ with positive direction of $X$-axis in anti-clock wise direction.
${m_2}$ be the slope of tangent ${PT_2}$ which makes an angle ${\theta _2}$ with positive direction of $X$-axis in anti-clock wise direction.
Using the concept of tangents explained in previous section, we can write $${m_1} = \tan {\theta _1} = {\left( {\frac{{dy}}{{dx}}} \right)_{{c_1}}}$$ and $${m_2} = \tan {\theta _2} = {\left( {\frac{{dy}}{{dx}}} \right)_{{c_2}}}$$ From figure, In $\Delta P{T_1}{T_2}$$$\angle P{T_1}{T_2} = {\theta _1},\quad \angle P{T_2}{T_1} = {180^ \circ } - {\theta _2},\quad \angle {T_1}P{T_2} = \theta $$Therefore, Sum of all the angles in a triangle is ${180^ \circ }$, we can write $$\begin{equation} \begin{aligned} {\theta _1} + {180^ \circ } - {\theta _2} + \theta = {180^ \circ } \\ \theta = {\theta _2} - {\theta _1}\quad ...(1) \\\end{aligned} \end{equation} $$ Using trigonometry, we can write
$$\begin{equation} \begin{aligned} \tan \theta = \tan ({\theta _2} - {\theta _1}) = \frac{{\tan {\theta _2} - \tan {\theta _1}}}{{1 + \tan {\theta _2}\tan {\theta _1}}} \\ \tan \theta = \left| {\frac{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_2}}} - {{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_1}}}}}{{1 + {{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_2}}}{{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_1}}}}}} \right| \\\end{aligned} \end{equation} $$
Note:
1. If angle of intersection of two curve is ${90^ \circ }$ i.e., $\theta = {90^ \circ }$
$$\begin{equation} \begin{aligned} \tan ({90^ \circ }) = \left| {\frac{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_2}}} - {{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_1}}}}}{{1 + {{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_2}}}{{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_1}}}}}} \right| = \infty \\ \Rightarrow 1 + {\left( {\frac{{dy}}{{dx}}} \right)_{{c_2}}}{\left( {\frac{{dy}}{{dx}}} \right)_{{c_1}}} = 0 \\ \therefore {\left( {\frac{{dy}}{{dx}}} \right)_{{c_2}}}{\left( {\frac{{dy}}{{dx}}} \right)_{{c_1}}} = - 1 \\ or \\ {m_1}.{m_2} = - 1 \\\end{aligned} \end{equation} $$ Curves satisfying this condition are called Orthogonal Curves.
2. If angle of intersection of two curve is ${0^ \circ }$ i.e., $\theta = {0^ \circ }$
$$\begin{equation} \begin{aligned} \tan ({0^ \circ }) = \left| {\frac{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_2}}} - {{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_1}}}}}{{1 + {{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_2}}}{{\left( {\frac{{dy}}{{dx}}} \right)}_{{c_1}}}}}} \right| = 0 \\ \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{{c_2}}} - {\left( {\frac{{dy}}{{dx}}} \right)_{{c_1}}} = 0 \\ \therefore {\left( {\frac{{dy}}{{dx}}} \right)_{{c_2}}} = {\left( {\frac{{dy}}{{dx}}} \right)_{{c_1}}} \\ or \\ {m_1} = {m_2} \\\end{aligned} \end{equation} $$
Example 4. Find the angle of intersection of two curves ${y^2} = 4x$ and ${x^2} = 4y$.
Solution: To find the angle of intersection, first of all we need the co-ordinates of point of intersection of two curves. The given equation of curves are $$\begin{equation} \begin{aligned} {x^2} = 4y\quad ...(1) \\ {y^2} = 4x\quad ...(2) \\\end{aligned} \end{equation} $$ Put the value of $x$ from equation $(2)$ in equation $(1)$, we get $$\begin{equation} \begin{aligned} {\left( {\frac{{{y^2}}}{4}} \right)^2} = 4y \\ {y^4} - 64y = 0 \\ y({y^3} - 64) = 0 \\ y = 0,\;y = 4 \\ \Rightarrow x = 0,\;x = 4 \\\end{aligned} \end{equation} $$ Therefore, the point of intersection of two curves are $(0,0)$ and $(4,4)$. To find the angle between two curves, we have to differentiate the equation of curves to find the slope of tangents drawn at the point of intersection of curves i.e., $$\begin{equation} \begin{aligned} {x^2} = 4y\quad ...(1) \\ \Rightarrow 2x = 4\frac{{dy}}{{dx}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{x}{2} = {m_1} \\ {y^2} = 4x\quad ...(2) \\ \Rightarrow 2y\frac{{dy}}{{dx}} = 4 \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{y} = {m_2} \\\end{aligned} \end{equation} $$ Now, the angle of intersection of two curves is $$\begin{equation} \begin{aligned} \tan \theta = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right| \\ {\text{ }} = \left| {\frac{{\frac{2}{y} - \frac{x}{2}}}{{1 + \frac{2}{y} \times \frac{x}{2}}}} \right| = \left| {\frac{{16 - {y^2}{x^2}}}{{2{x^2} + 2{y^2}}}} \right| \\\end{aligned} \end{equation} $$ Put the values of co-ordinates of point of intersection of curves i.e., $(0,0)$ and $(4,4)$, we get $$\begin{equation} \begin{aligned} \tan {\theta _{(0,0)}} = \left| {\frac{{16 - {0^2}{0^2}}}{{{{2.0}^2} + {{2.0}^2}}}} \right| = \infty \\ {\theta _{(0,0)}} = \frac{\pi }{2} \\ \tan {\theta _{(4,4)}} = \left| {\frac{{16 - {4^2}{4^2}}}{{{{2.4}^2} + {{2.4}^2}}}} \right| = \left| {\frac{3}{4}} \right| \\ {\theta _{(4,4)}} = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right) \\\end{aligned} \end{equation} $$
Example 5. If the curves $x = {y^2}$ and $xy=k$ are orthogonal then find the value of $k$.
Solution: As we know if the two curves are orthogonal, the angle of intersection of them is ${90^ \circ }$. First we differentiate both the equation of curves with respect to $x$ to find the slope of tangent i.e., $$\begin{equation} \begin{aligned} x = {y^2}\quad ...(1) \\ \Rightarrow 1 = 2y\frac{{dy}}{{dx}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2y}} = {m_1} \\ xy = k\quad ...(2) \\ \Rightarrow x\frac{{dy}}{{dx}} + y = 0 \\ \Rightarrow \frac{{dy}}{{dx}} = - \frac{y}{x} = {m_2} \\\end{aligned} \end{equation} $$ Now, we have to find the point of intersection of two curves i.e., by putting the value of $x$ from equation $(1)$ in equation $(2)$, we get $$\begin{equation} \begin{aligned} {y^2}.y = k \\ \Rightarrow {y^3} = k \\ \Rightarrow y = {k^{\frac{1}{3}}} \\ \therefore x = {k^{\frac{2}{3}}} \\\end{aligned} \end{equation} $$ The point of intersection of two curves is $$\left( {{k^{\frac{2}{3}}},{k^{\frac{1}{3}}}} \right)$$ Now, the angle of intersection of two curves is given by $$\begin{equation} \begin{aligned} \tan \theta = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_2}{m_1}}}} \right| \\ \tan {90^ \circ } = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_2}{m_1}}}} \right| = \infty \\ \Rightarrow 1 + {m_2}{m_1} = 0 \\ \Rightarrow {m_2}{m_1} = - 1 \\\end{aligned} \end{equation} $$ Put the value of ${m_1}$ and ${m_2}$ we get, $$\begin{equation} \begin{aligned} \frac{1}{{2y}} \times - \frac{y}{x} = - 1 \\ \frac{1}{{2x}} = 1 \\\end{aligned} \end{equation} $$ Now, put the value of point of intersection in it, we get $$\begin{equation} \begin{aligned} \frac{1}{{2 \times {k^{\frac{2}{3}}}}} = 1 \\ \Rightarrow {k^{\frac{2}{3}}} = \frac{1}{2} \\\end{aligned} \end{equation} $$ To find the value of ${k^2}$, take cube both sides, we get $$\begin{equation} \begin{aligned} {k^{\frac{2}{3} \times 3}} = {\left( {\frac{1}{2}} \right)^3} \\ {k^2} = \frac{1}{8} \\\end{aligned} \end{equation} $$
Example 6. If two curves $a{x^2} + b{y^2} = 1$ and $a'{x^2} + b'{y^2} = 1$ intersect orthogonally, find the relation between $a$, $b$, $a'$ and $b'$.
Solution: As we know if the two curves are orthogonal, the angle of intersection of them is ${90^ \circ }$. First we differentiate both the equation of curves with respect to $x$ to find the slope of tangent i.e., $$\begin{equation} \begin{aligned} a{x^2} + b{y^2} = 1\quad ...(1) \\ 2ax + 2by\frac{{dy}}{{dx}} = 0 \\ \frac{{dy}}{{dx}} = \frac{{ax}}{{by}} = {m_1} \\ a'{x^2} + b'{y^2} = 1\quad ...(2) \\ 2a'x + 2b'y\frac{{dy}}{{dx}} = 0 \\ \frac{{dy}}{{dx}} = \frac{{a'x}}{{b'y}} = {m_2} \\\end{aligned} \end{equation} $$ Now, we have to find the point of intersection of two curves i.e., $$\begin{equation} \begin{aligned} a{x^2} + b{y^2} = a'{x^2} + b'{y^2} \\ {x^2}(a - a') = {y^2}(b - b') \\ \frac{{{x^2}}}{{{y^2}}} = \frac{{b' - b}}{{a - a'}}\quad ...(3) \\\end{aligned} \end{equation} $$ Now, the angle of intersection of two curves is given by $$\begin{equation} \begin{aligned} \tan \theta = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_2}{m_1}}}} \right| \\ \tan {90^ \circ } = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_2}{m_1}}}} \right| = \infty \\ \Rightarrow 1 + {m_2}{m_1} = 0 \\ \Rightarrow {m_2}{m_1} = - 1 \\\end{aligned} \end{equation} $$ Put the value of ${m_1}$ and ${m_2}$ we get, $$\begin{equation} \begin{aligned} \frac{{ax}}{{by}} \times \frac{{a'x}}{{b'y}} = - 1 \\ \frac{{aa'}}{{bb'}} \times \frac{{{x^2}}}{{{y^2}}} = - 1 \\\end{aligned} \end{equation} $$ Put the value from equation $(3)$, we get $$\begin{equation} \begin{aligned} \frac{{aa'}}{{bb'}} \times \frac{{b' - b}}{{a - a'}} = - 1 \\ aa'b' - aa'b = - bb'a + bb'a' \\\end{aligned} \end{equation} $$ Divide by $aa'bb'$, we get $$\begin{equation} \begin{aligned} \frac{{aa'b'}}{{aa'bb'}} - \frac{{aa'b}}{{aa'bb'}} = - \frac{{bb'a}}{{aa'bb'}} + \frac{{bb'a'}}{{aa'bb'}} \\ \frac{1}{b} - \frac{1}{{b'}} = - \frac{1}{{a'}} + \frac{1}{a} \\ \frac{1}{a} - \frac{1}{b} = \frac{1}{{a'}} - \frac{1}{{b'}} \\\end{aligned} \end{equation} $$
Note: Using the above derived results, we can conclude that
- If two ellipse $\frac{{{x^2}}}{{{a_1}^2}} + \frac{{{y^2}}}{{{b_1}^2}} = 1$ and $\frac{{{x^2}}}{{{a_2}^2}} + \frac{{{y^2}}}{{{b_2}^2}} = 1$ intersect orthogonally, then $${a_1}^2 - {b_1}^2 = {a_2}^2 - {b_2}^2$$
- If two hyperbola $\frac{{{x^2}}}{{{a_1}^2}} - \frac{{{y^2}}}{{{b_1}^2}} = 1$ and $\frac{{{x^2}}}{{{a_2}^2}} - \frac{{{y^2}}}{{{b_2}^2}} = 1$ intersect orthogonally, then $${a_1}^2 + {b_1}^2 = {a_2}^2 + {b_2}^2$$
- If ellipse $\frac{{{x^2}}}{{{a_1}^2}} + \frac{{{y^2}}}{{{b_1}^2}} = 1$ and hyperbola $\frac{{{x^2}}}{{{a_2}^2}} - \frac{{{y^2}}}{{{b_2}^2}} = 1$ intersect orthogonally, then $${a_1}^2 - {b_1}^2 = {a_2}^2 + {b_2}^2$$
