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Question: The eccentric angle of point of intersection of the ellipse \[{{x}^{2}}+4{{y}^{2}}=4\] and the parab...

The eccentric angle of point of intersection of the ellipse x2+4y2=4{{x}^{2}}+4{{y}^{2}}=4 and the parabola x2+1=y{{x}^{2}}+1=y is;
(A) 00
(B) cos⁡−1(−23){{\cos }^{-1}}\left( -\dfrac{2}{3} \right)
(C) π2\dfrac{\pi }{2}
(D) 5π4\dfrac{5\pi }{4}

Explanation

Solution

Hint: Consider a variable parametric point on the ellipse and substitute the same point on the given parabola as both the curves intersect, to find out the eccentric angle.

Complete step-by-step answer:
The given ellipse equation x2+4y2=4{{x}^{2}}+4{{y}^{2}}=4, can be rewritten as:
x24+y21=1\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{1}=1
As we know, for any given eccentricity ‘θ\theta ’, a variable point on ellipse x2a2+y2b2=1\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 can be considered as (acos⁡θ,bsin⁡θ)\left( a\cos \theta ,b\sin \theta \right), now for x24+y21=1\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{1}=1, the variable point on the ellipse will be (2cos⁡θ,sin⁡θ)\left( 2\cos \theta ,\sin \theta \right).
As the ellipse intersects the parabola x2+1=y{{x}^{2}}+1=y.
The point (2cos⁡θ,sin⁡θ)\left( 2\cos \theta ,\sin \theta \right) thet we considered will also lie on that parabola.
So now substitute the point in x2+1=y{{x}^{2}}+1=y, which is the given parabola equation.
We have:
(2cos⁡θ)2+1=sin⁡θ{{\left( 2\cos \theta \right)}^{2}}+1=\sin \theta
4cos⁡2θ+1=sin⁡θ4{{\cos }^{2}}\theta +1=\sin \theta
Now, applying the trigonometry identity cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta , we will have:
4(1−sin⁡2θ)+1=sin⁡θ4\left( 1-{{\sin }^{2}}\theta \right)+1=\sin \theta
4sin⁡2θ+sin⁡θ−5=04{{\sin }^{2}}\theta +\sin \theta -5=0
4sin⁡2θ+5sin⁡θ−4sin⁡θ−5=04{{\sin }^{2}}\theta +5\sin \theta -4\sin \theta -5=0
Factoring the above equation, we will have:
(4sin⁡θ+5)(sin⁡θ−1)=0\left( 4\sin \theta +5 \right)\left( \sin \theta -1 \right)=0
sin⁡θ=−54\sin \theta =-\dfrac{5}{4} (or) sin⁡θ=1\sin \theta =1
So, the eccentricity is θ=sin⁡−1=π2\theta ={{\sin }^{-1}}=\dfrac{\pi }{2}.
As sin⁡θ=−54\sin \theta =-\dfrac{5}{4} is not possible, as it does not lie within the range of the function.
The range of sin⁡θ\sin \theta and cos⁡θ\cos \theta functions is [−1,1][-1,1] only, so keep this in mind while solving trigonometric equations.
So, the eccentricity is π2\dfrac{\pi }{2}.
Hence, option c is the correct answer.
Note: The range of sin⁡θ\sin \theta and cos⁡θ\cos \theta functions is [−1,1][-1,1] only, so keep this in mind while solving trigonometric equations.