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Question: If the parabola \({{y}^{2}}=4x\) and \({{x}^{2}}=32y\) intersect at (16, 8) at an angle \(\theta \) ...

If the parabola y2=4x{{y}^{2}}=4x and x2=32y{{x}^{2}}=32y intersect at (16, 8) at an angle θ\theta .

& a){{\tan }^{-1}}\left( \dfrac{3}{5} \right) \\\ & b){{\tan }^{-1}}\left( \dfrac{4}{5} \right) \\\ & c)\pi \\\ & d)\dfrac{\pi }{2} \\\ \end{aligned}$$
Explanation

Solution

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 tan1(m2m11+m1m2)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 y2=4x{{y}^{2}}=4x and x2=32y{{x}^{2}}=32y
First consider the parabola y2=4x{{y}^{2}}=4x
Now let us differentiate the equation on both sides with respect to x and hence we get.
2ydydx=42y\dfrac{dy}{dx}=4
Now rearranging the terms we get
dydx=42y\dfrac{dy}{dx}=\dfrac{4}{2y}
Now we know that this is nothing but slope of tangent to parabola y2=4x{{y}^{2}}=4x at point (x, y)
Now Let us calculate the slope of tangent at (16, 8)
dydx=42×8=14\dfrac{dy}{dx}=\dfrac{4}{2\times 8}=\dfrac{1}{4}
Let us call this slope as m1{{m}_{1}}
Hence we have the slope of tangent of parabola y2=4x{{y}^{2}}=4x at (16, 8) is m1=14{{m}_{1}}=\dfrac{1}{4}………………. (1)
Now consider the second parabola x2=32y{{x}^{2}}=32y
Differentiating this equation on both sides we get
2x=32dydx2x=32\dfrac{dy}{dx}
Rearranging the terms we get
dydx=2x32\dfrac{dy}{dx}=\dfrac{2x}{32}
Now we know that this is nothing but slope of tangent to parabola x2=32y{{x}^{2}}=32y at point (x, y)
Now Let us calculate the slope of tangent at (16, 8)
dydx=2×1632=1\dfrac{dy}{dx}=\dfrac{2\times 16}{32}=1
Now let us call this slope as m2{{m}_{2}}
Hence we have the slope of tangent to parabola x2=32y{{x}^{2}}=32y at (16, 8) is m2=1{{m}_{2}}=1 ……………….. (2)
Now we know that the angle between two lines is with slope m1{{m}_{1}} and m2{{m}_{2}} is given by
θ=tan1(m2m11+m1m2)\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
θ=tan1(1141+(1)(14)) θ=tan1(4144+14) θ=tan1(35) \begin{aligned} & \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{aligned}
Hence we have the value of θ\theta is tan1(35){{\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.