Question

Find the angle between the curves $xy=2$ and $x^2+4y=0$.

Answer 1

Hint:Calculate the point of intersection of two curves. Calculate the slope of the tangent of both the curves at their point of intersection by finding the value of the first derivative of the curve at the point of intersection. Use the formula $\tan \theta =\left|{\displaystyle \frac{m_1-m_2}{1+m_1{m}_2}}\right|$ to calculate the angle between the tangents at point of intersection, where $m_1$ and $m_2$ are slopes of tangents.

Complete step-by-step answer:
We have to calculate the angles between the curves $xy=2$ and $x^2+4y=0$.
 

We will first find the point of intersection of the two curves.
We know that $xy=2$. Rearranging the terms of the above equation, we have $y={\displaystyle \frac{2}{x}}$. Substituting this equation in the equation $x^2+4y=0$, we have $x^2+4\left({\displaystyle \frac{2}{x}}\right)=0$.
Thus, we have ${\displaystyle \frac{x^3+8}{x}}=0\Rightarrow x^3+8=0$.
So, we have $x^3=-8={(-2)}^3$. Taking the cube root on both sides, we have $x=-2.....\left(1\right)$.
Substituting equation (1) in the equation $y={\displaystyle \frac{2}{x}}$, we have $y={\displaystyle \frac{2}{-2}}=-1$.
Thus, the point of intersection of two curves is $A=(-2,-1)$.
We know that the angle between the two curves is the angle between the tangents of both the curves at their point of intersection.
We will now calculate the slope of tangents of both the curves at their point of intersection. To do so, we will calculate the value of the first derivative of both the curves at the point of intersection.
We will first consider the curve $xy=2\Rightarrow y={\displaystyle \frac{2}{x}}$.
We know that differentiation of any function of the form $y=a{x}^n$ is ${\displaystyle \frac{dy}{dx}}=an{x}^{n-1}$.
Thus, for $y={\displaystyle \frac{2}{x}}$, we have ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-2}{x^2}}$. Substituting the point $A=(-2,-1)$in the previous equation, we have ${\left({\displaystyle \frac{dy}{dx}}\right)}_{x=-2}={\displaystyle \frac{-2}{{(-2)}^2}}={\displaystyle \frac{-1}{2}}$.
We will now calculate the slope of the tangent of the curve $x^2+4y=0\Rightarrow y={\displaystyle \frac{-x^2}{4}}$.
Thus, we have ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-2x}{4}}={\displaystyle \frac{-x}{2}}$. Substituting the point $A=(-2,-1)$in the previous equation, we have ${\left({\displaystyle \frac{dy}{dx}}\right)}_{x=-2}={\displaystyle \frac{-(-2)}{2}}=1$.
Thus, the slope of tangents of the curves $xy=2$ and $x^2+4y=0$ at their point of intersection is ${\displaystyle \frac{-1}{2}}$ and $1$.
We will now calculate the angle between the tangents at the point of intersection of the two curves.
We know that angle between two tangents with slopes $m_1$ and $m_2$ is given by $\tan \theta =\left|{\displaystyle \frac{m_1-m_2}{1+m_1{m}_2}}\right|$.
Substituting $m_1$=${\displaystyle \frac{-1}{2}}$,$m_2=1$ in the above formula, we have $\tan \theta =\left|{\displaystyle \frac{{\displaystyle \frac{-1}{2}}-\left(1\right)}{1+\left(1\right)\left({\displaystyle \frac{-1}{2}}\right)}}\right|$.
Simplifying the above expression, we have $\tan \theta =\left|{\displaystyle \frac{{\displaystyle \frac{-1}{2}}-\left(1\right)}{1+\left(1\right)\left({\displaystyle \frac{-1}{2}}\right)}}\right|=\left|{\displaystyle \frac{{\displaystyle \frac{-1}{2}}-1}{1+{\displaystyle \frac{1}{2}}}}\right|=\left|{\displaystyle \frac{{\displaystyle \frac{-3}{2}}}{{\displaystyle \frac{1}{2}}}}\right|=3$.
Taking inverse on both sides, we have $\theta ={\tan }^{-1}\left(3\right)$.
Hence, the angle between the curves $xy=2$ and $x^2+4y=0$ is $\theta ={\tan }^{-1}\left(3\right)$.

Note: We can’t solve this question without using the fact that the angle between the two curves is the angle between the tangents of both the curves at their point of intersection. We can also solve this question by writing the exact equation of tangents at the point of intersection of two curves and then calculating the angles between them.