Angle of Intersection between Two Curves: Formula, Examples

The angle of Intersection of Two Curves

Let y = f (x) and y = g (x) be two curves intersecting at a point P(x₀, y₀) . Then the angle of intersection of two curves is defined as the angle between the tangent to the two curves at the point of intersection.

Let $P T_1$ and $P T_2$ be tangents to the curve $y=f(x)$ and $y=g(x)$ at their point of intersection.
Let $\Theta$ be the angle between two tangents $P T_1$ and $P T_2, \Theta_1$ and $\Theta_2$ are angles made by tangents $P T_1$ and $P T_2$ with the positive direction of $x$-axis, then

$
\begin{aligned}
& m_1=\tan \theta_1=\left(\frac{d}{d x}(f(x))\right)_{\left(x 0, y_0\right)} \\
& m_2=\tan \theta_2=\left(\frac{d}{d x}(g(x))\right)_{\left(x_0, y_0\right)}
\end{aligned}
$

from the figure $\theta=\theta_1-\theta_2$

$
\begin{aligned}
& \Rightarrow \tan \theta=\tan \left(\theta_1-\theta_2\right)=\frac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_1 \cdot \tan \theta_2} \\
& \Rightarrow \tan \theta=\left|\frac{\left(\frac{\mathbf{d}}{\mathrm{dx}}(\mathbf{f}(\mathbf{x}))\right)_{\left(\mathbf{x}_0, \mathbf{y}_{\mathbf{o}}\right)}-\left(\frac{\mathrm{d}}{\mathrm{dx}}(\mathbf{g}(\mathbf{x}))\right)_{\left(\mathbf{x}_0, \mathbf{y}_{\mathbf{o}}\right)}}{1+\left(\frac{\mathrm{d}}{\mathrm{dx}}(\mathbf{f}(\mathbf{x}))\right)_{\left(\mathbf{x}_{\mathbf{0}}, \mathbf{y}_{\mathbf{o}}\right)} \cdot\left(\frac{\mathrm{d}}{\mathbf{d x}}(\mathbf{g}(\mathbf{x}))\right)_{\left(\mathbf{x}_0, \mathbf{y}_0\right)}}\right|
\end{aligned}
$

Orthogonal Curves

If the angle of the intersection of two curves is a right angle then two curves are called orthogonal curves.
In this case, $\tan \theta=90^{\circ}$

$
\Rightarrow\left(\frac{d}{d x}(f(x))\right)_{\left(x_0, y_0\right)} \cdot\left(\frac{d}{d x}(g(x))\right)_{\left(x_0, y_0\right)}=-1
$

this is also the condition for two curves to be orthogonal.

Condition for two curves to touch each other

$
\left(\frac{d}{d x}(f(x))\right)_{\left(x_0, y_0\right)}=\left(\frac{d}{d x}(g(x))\right)_{\left(x_0, y_0\right)}
$

Recommended Video Based on Angle of Intersection of Two Curves:

Solved Examples Based On Angle of Intersection of Two Curves:

Example 1: If the curves $y^2=6 x, 9 x^2+b y^2=16$ intersect each other at right angles, then the value of b is :
[JEE Main 2028]
1) $9 / 2$
2) 6
3) $7 / 2$
4) 4

Solution
As we have learned
Condition of Orthogonality -
Two curves intersect each other orthogonally if the tangents to each of them subtend a right angle at the point of intersection of two curves:

$
m_1 \times m_2=-1
$

$
\begin{aligned}
& y^2=6 x \ldots \ldots \ldots \\
& 9 x^2+6 b x=16
\end{aligned}
$

$\qquad$ $\qquad$
Slope of tangent of first curve

$
\begin{aligned}
& 2 y \frac{d y}{d x}=6 \Rightarrow \frac{d y}{d x}=\frac{6}{2 y} \\
& m_1=\frac{6}{2 y}
\end{aligned}
$

Slope of tangent of second curve

$
\begin{aligned}
& 18 x+2 b y \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{-9 x}{b y} \\
& \frac{-9 x}{b y}=m_2
\end{aligned}
$

So $m_1 \times m_2=-1$

$
\begin{aligned}
& m_1 m_2=-1 \\
& \left(\frac{6}{2 y}\right)\left(\frac{-b x}{9 y}\right)=1 \\
& -27 x=-b y^2 \\
& -27 x=-b(6 x) \\
& b=\frac{27}{6}=\frac{9}{2}
\end{aligned}
$

Example 2: $\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$
[JEE Main 2019]
1) exists and equals $\frac{1}{4 \sqrt{2}}$
2) exists and equals $\frac{1}{2 \sqrt{2}(\sqrt{2}+1)}$
3) exists and equals $\frac{1}{2 \sqrt{2}}$
4) does not exist|

Solution
Angle of intersection of two curves -
The angle of intersection of two curves is the angle subtended between the tangents at their point of intersection Let $m_1$ \& $m_2$ are two slope of tangents at intersection point of two curves then

$
\tan \theta=\frac{\left[m_1-m_2\right]}{1+m_1 m_2}
$

where $\theta$ is angle between two curves tangents.

$
\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+r^4}}-\sqrt{2}}{y^4}
$

$
\begin{aligned}
& \Rightarrow \lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} \times \frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}} \\
& \Rightarrow \lim _{y \rightarrow 0} \frac{1+\sqrt{1+y^4}-2}{y^4\left(\sqrt{1+\sqrt{1+y^4}}\right)+\sqrt{2}}
\end{aligned}
$

again factorize

$
\begin{aligned}
& \Rightarrow \lim _{y \rightarrow 0} \frac{\left(\sqrt{1+y^4}\right)-1}{y^4\left(\sqrt{1+\sqrt{1+y^4}}\right)+\sqrt{2}} \times \frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1} \\
& \Rightarrow \lim _{y \rightarrow 0} \frac{1+y^4-1}{y^4\left(\sqrt{1+\sqrt{1+y^4}+\sqrt{2}}\right)\left(\sqrt{1+y^4}+1\right)} \\
& \Rightarrow \lim _{y \rightarrow 0} \frac{1}{\left(\sqrt{1+\sqrt{1+y^4}+\sqrt{2}}\right)\left(\sqrt{1+y^4}+1\right)}=\frac{1}{4 \sqrt{2}}
\end{aligned}
$

ellipse $\frac{x^2}{9}+\frac{y^2}{1}=1$ and the circle $x^2+y^2=3$

Example 3: Let $\theta$ be the acute angle between the tangents to the at their point of intersection in the first quadrant. Then $\tan \Theta$ is equal to:
[JEE Main 2021]
1) $\frac{5}{2 \sqrt{3}}$
2) $\frac{4}{\sqrt{3}}$
3) $\frac{2}{\sqrt{3}}$
4) 2

Solution

$
\begin{aligned}
& x^2=3-y^2, \frac{x^2}{9}+y^2=1 \\
& \Rightarrow 3-y^2+3 y^2=9 \Rightarrow 8 y^2=6 \\
& \Rightarrow x^2=3-\frac{3}{4}=\frac{9}{4} \Rightarrow x= \pm \frac{3}{2}
\end{aligned}
$

So, the point of intersection in the first quadrant is

$
\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)
$
Slope of Tangent to ellipse

$
=m_1: \frac{-1}{9} \frac{x_1}{y_1}=\frac{-\sqrt{3}}{9} \text { slope of Tangent to circle }=m_2: \frac{-x_1}{y_1}=-\sqrt{3}
$

$\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\sqrt{3}-\frac{\sqrt{3}}{9}}{1+\frac{3}{9}}\right|$

$
=\frac{8 \sqrt{3}}{12}=\frac{2 \sqrt{3}}{3}=\frac{2}{\sqrt{3}}
$
Hence, the answer is the option (3).

Example 4: An angle of intersection of the curves $\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$ and $x^2+y^2=\mathrm{ab}, \mathrm{a}>\mathrm{b}$ is :
[JEE Main 2021]

$
\begin{aligned}
& \text { 1) } \tan ^{-1}(2 \sqrt{a b}) \\
& \tan ^{-1}\left(\frac{a+b}{\sqrt{a b}}\right) \\
& \text { 2) } \tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right) \\
& \text { 4) } \tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)
\end{aligned}
$

Solution

Find a point of intersection of the curves

$
\begin{aligned}
& x^2=a b-y^2 \\
& \Rightarrow \frac{a b-y^2}{a^2}+\frac{y^2}{b^2}=1 \\
& \Rightarrow y^2\left(\frac{1}{b^2-\frac{1}{a^2}}\right)=1-\frac{a b}{a^2}=1-\frac{b}{a} \\
& \Rightarrow y^2 \frac{\left(a^2-b^2\right)}{a^2 b^2}=\frac{a-b}{a} \\
& \Rightarrow y^2=\frac{a b^2}{a+b} \Rightarrow y=b \sqrt{\frac{a}{a+b}} \\
& x^2=a b-\frac{a b^2}{a+b}=\frac{a^2 b}{a+b} \Rightarrow x=a \sqrt{\frac{b}{a+b}} \\
& \text { So, }\left(x_1, y_1\right)=\left(a \sqrt{\frac{b}{a+b}}, b \sqrt{\frac{a}{a+b}}\right) \\
& C_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\end{aligned}
$

Slope of tangent at $\left(x_1, y_1\right)$

$
\begin{aligned}
& m_1=\frac{-x_1 b^2}{y_1 a^2}=\frac{-a \sqrt{b}}{b \sqrt{a}} \times \frac{b^2}{a^2}=-\left(\frac{b}{a}\right)^{3 / 2} \\
& C_2: x^2-y^2=a b
\end{aligned}
$

Slope of tangent at $\left(x_1, y_1\right)$

$
\begin{aligned}
& m_2=\frac{-x_1}{y_1}=\frac{-a \sqrt{b}}{b \sqrt{a}}=-\left(\frac{a}{b}^{1 / 2}\right) \\
& \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|:\left|\frac{-\frac{b \sqrt{b}}{a \sqrt{a}}+\frac{\sqrt{a}}{\sqrt{b}}}{1+\frac{b \sqrt{b}}{a \sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{b}}}\right|=\left|\frac{\frac{a^2-b^2}{a \sqrt{a b}}}{\frac{a+b}{a}}\right| \\
& \Rightarrow \theta=\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)
\end{aligned}
$

Hence, the answer is the option (2).

Example 3: $\frac{x^2}{\alpha}+\frac{y^2}{4}=1$ and $y^3=16 x$ intersect at right angles, then a value of $\alpha$ is:
[JEE Main 2013]
1) 2
2) $\frac{4}{3}$
3) $\frac{1}{2}$
4) $\frac{3}{4}$

Solution

$
\begin{aligned}
& \frac{x^2}{\alpha}+\frac{y^2}{4}=1 \Rightarrow \frac{2 x}{\alpha}+\frac{2 y}{4} \cdot \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{-4 x}{\alpha y} \\
& y^3=16 x \\
& \Rightarrow 3 y^2 \frac{d y}{d x}=16 \\
& \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2}
\end{aligned}
$

Since, the curves intersect at right angles, then

$
\begin{aligned}
& \therefore \frac{-4 x}{\alpha y} \times \frac{16}{3 y^2}=-1 \Rightarrow 3 \alpha y^3=64 x \\
& \Rightarrow \alpha=\frac{64 x}{3 \times 16 x}=\frac{4}{3}
\end{aligned}
$

Hence, the answer is the option (2).