Question

If the parabola ${{y}^{2}}=4x$ and ${{x}^{2}}=32y$ intersect at (16, 8) at an angle $\theta $ .
\[\begin{align}
  & a){{\tan }^{-1}}\left( \dfrac{3}{5} \right) \\
 & b){{\tan }^{-1}}\left( \dfrac{4}{5} \right) \\
 & c)\pi \\
 & d)\dfrac{\pi }{2} \\
\end{align}\]

Answer 1

Hint: Now we know the equation of parabolas. On differentiating both parabolas we will get the slope of both parabolas at any point (x, y). Now we will calculate the slope of tangents at point (16, 8). Now once we have slope of both tangents we know that the angle between two lines is given by $ta{{n}^{-1}}\left( \dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right)$ .

Complete step by step answer:
Now we are given with two parabolas ${{y}^{2}}=4x$ and ${{x}^{2}}=32y$
First consider the parabola ${{y}^{2}}=4x$
Now let us differentiate the equation on both sides with respect to x and hence we get.
$2y\dfrac{dy}{dx}=4$
Now rearranging the terms we get
$\dfrac{dy}{dx}=\dfrac{4}{2y}$
Now we know that this is nothing but slope of tangent to parabola ${{y}^{2}}=4x$ at point (x, y)
Now Let us calculate the slope of tangent at (16, 8)
$\dfrac{dy}{dx}=\dfrac{4}{2\times 8}=\dfrac{1}{4}$
Let us call this slope as ${{m}_{1}}$
Hence we have the slope of tangent of parabola ${{y}^{2}}=4x$ at (16, 8) is ${{m}_{1}}=\dfrac{1}{4}$………………. (1)
Now consider the second parabola ${{x}^{2}}=32y$
Differentiating this equation on both sides we get
$2x=32\dfrac{dy}{dx}$
Rearranging the terms we get
$\dfrac{dy}{dx}=\dfrac{2x}{32}$
Now we know that this is nothing but slope of tangent to parabola ${{x}^{2}}=32y$ at point (x, y)
Now Let us calculate the slope of tangent at (16, 8)
$\dfrac{dy}{dx}=\dfrac{2\times 16}{32}=1$
Now let us call this slope as ${{m}_{2}}$
Hence we have the slope of tangent to parabola ${{x}^{2}}=32y$ at (16, 8) is ${{m}_{2}}=1$ ……………….. (2)
Now we know that the angle between two lines is with slope ${{m}_{1}}$ and ${{m}_{2}}$ is given by
$\theta ={{\tan }^{-1}}\left( \dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right)$
Hence we have from equation (1) and equation (2) that
$\begin{align}
  & \theta ={{\tan }^{-1}}\left( \dfrac{1-\dfrac{1}{4}}{1+\left( 1 \right)\left( \dfrac{1}{4} \right)} \right) \\
 & \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{4-1}{4}}{\dfrac{4+1}{4}} \right) \\
 & \theta ={{\tan }^{-1}}\left( \dfrac{3}{5} \right) \\
\end{align}$
Hence we have the value of $\theta $ is ${{\tan }^{-1}}\left( \dfrac{3}{5} \right)$

So, the correct answer is “Option A”.

Note: Now note that be it any curve the angle between two curves is given by the angle between their tangents at that point. Hence when we are asked to find the angle between the curves at a point we have to find the angle between their tangents at that point. We can also find angles between their normal at the point.