Question

The angle between the curves $y^2=4x+4$ and $y^2=36\left(9-x\right)$ is

• A. $30°$
• B. $45°$
• C. $60°$
• D. $90°$

Answer 1

The correct option is D

$90°$

Explanation for the correct option:

Step 1: Finding the values of $m_1$ and $m_2$

Given, the curves are

$y^2=4x+4......\left(1\right)$

$y^2=36\left(9-x\right)......\left(2\right)$

Differentiating Equation $\left(1\right)$ with respect to $x$

$$\begin{gathered} 2y\frac{dy}{dx}=4 \\ \Rightarrow \frac{dy}{dx}=\frac{4}{2y} \\ \Rightarrow m_1=\frac{2}{y}......\left(3\right) \end{gathered}$$

Differentiating Equation $\left(2\right)$ with respect to $x$

$$\begin{gathered} 2y\frac{dy}{dx}=-36 \\ \Rightarrow \frac{dy}{dx}=-\frac{36}{2y} \\ \Rightarrow m_2=-\frac{18}{y}......\left(4\right) \end{gathered}$$

Step 2: Finding the angle between the curves

We know that angle between the tangents is given by,

$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

Substituting the value of $m_1$and $m_2$ in the above formula we get,

$\begin{array}{rcl}\tan \theta & =& \left|\frac{-{\displaystyle \frac{18}{y}}-{\displaystyle \frac{2}{y}}}{1+\left(-{\displaystyle \frac{18}{y}}\right)\left({\displaystyle \frac{2}{y}}\right)}\right|\\ & =& \left|\frac{{\displaystyle -\frac{20}{y}}}{1-{\displaystyle \frac{36}{y^2}}}\right|\\ & =& \frac{{\displaystyle \frac{20}{y}}}{{\displaystyle \frac{y^2-36}{y^2}}}\\ & =& \frac{20}{y}\times \frac{y^2}{y^2-36}\\ & =& \frac{20y}{y^2-36}......\left(5\right)\end{array}$

Also, we can write:

$$\begin{gathered} 4x+4=36\left(9-x\right) \\ \Rightarrow 4\left(x+1\right)=36\left(9-x\right) \\ \Rightarrow \left(x+1\right)=9\left(9-x\right) \\ \Rightarrow \left(x+1\right)=81-9x \\ \Rightarrow 10x=81-1 \\ \Rightarrow x=\frac{80}{10} \\ \Rightarrow x=8 \end{gathered}$$

Then, finding the value of $y$

$$\begin{gathered} y^2=4\left(8\right)+4 \\ \Rightarrow y^2=32+4 \\ \Rightarrow y^2=36 \\ \Rightarrow y=\sqrt{36} \\ \Rightarrow y=\pm 6 \end{gathered}$$

Substituting the value of $y$ in equation $\left(5\right)$ we get,

$$\begin{gathered} \Rightarrow \tan \theta =\frac{20\times 6}{6^2-36} \\ \Rightarrow \tan \theta =\frac{120}{36-36} \\ \Rightarrow \tan \theta =\frac{120}{0} \\ \Rightarrow \tan \theta =\infty \\ \Rightarrow \theta ={\tan }^{-1}\infty \\ \Rightarrow \theta =90°\mathbf{}\left[\because {\tan }^{-1}\infty =\frac{\pi }{2}=90°\right] \end{gathered}$$

Hence, option (D) is the correct answer.