The eccentric angle of point of intersection of the ellipse \[{{x}^{2}}+4{{y}^{2}}=4\] and the parabola \[{{x}^{2}}+1=y\] is;
(A) \[0\]
(B) \[{{\cos }^{-1}}\left( -\dfrac{2}{3} \right)\]
(C) \[\dfrac{\pi }{2}\]
(D) \[\dfrac{5\pi }{4}\]
Answer
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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 \[{{x}^{2}}+4{{y}^{2}}=4\], can be rewritten as:
\[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{1}=1\]
As we know, for any given eccentricity ‘\[\theta \]’, a variable point on ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] can be considered as \[\left( a\cos \theta ,b\sin \theta \right)\], now for \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{1}=1\], the variable point on the ellipse will be \[\left( 2\cos \theta ,\sin \theta \right)\].
As the ellipse intersects the parabola \[{{x}^{2}}+1=y\].
The point \[\left( 2\cos \theta ,\sin \theta \right)\] thet we considered will also lie on that parabola.
So now substitute the point in \[{{x}^{2}}+1=y\], which is the given parabola equation.
We have:
\[{{\left( 2\cos \theta \right)}^{2}}+1=\sin \theta \]
\[4{{\cos }^{2}}\theta +1=\sin \theta \]
Now, applying the trigonometry identity \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \], we will have:
\[4\left( 1-{{\sin }^{2}}\theta \right)+1=\sin \theta \]
\[4{{\sin }^{2}}\theta +\sin \theta -5=0\]
\[4{{\sin }^{2}}\theta +5\sin \theta -4\sin \theta -5=0\]
Factoring the above equation, we will have:
\[\left( 4\sin \theta +5 \right)\left( \sin \theta -1 \right)=0\]
\[\sin \theta =-\dfrac{5}{4}\] (or) \[\sin \theta =1\]
So, the eccentricity is \[\theta ={{\sin }^{-1}}=\dfrac{\pi }{2}\].
As \[\sin \theta =-\dfrac{5}{4}\] is not possible, as it does not lie within the range of the function.
The range of \[\sin \theta \] and \[\cos \theta \] functions is \[[-1,1]\] only, so keep this in mind while solving trigonometric equations.
So, the eccentricity is \[\dfrac{\pi }{2}\].
Hence, option c is the correct answer.
Note: The range of \[\sin \theta \] and \[\cos \theta \] functions is \[[-1,1]\] only, so keep this in mind while solving trigonometric equations.
Complete step-by-step answer:
The given ellipse equation \[{{x}^{2}}+4{{y}^{2}}=4\], can be rewritten as:
\[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{1}=1\]
As we know, for any given eccentricity ‘\[\theta \]’, a variable point on ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] can be considered as \[\left( a\cos \theta ,b\sin \theta \right)\], now for \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{1}=1\], the variable point on the ellipse will be \[\left( 2\cos \theta ,\sin \theta \right)\].
As the ellipse intersects the parabola \[{{x}^{2}}+1=y\].
The point \[\left( 2\cos \theta ,\sin \theta \right)\] thet we considered will also lie on that parabola.
So now substitute the point in \[{{x}^{2}}+1=y\], which is the given parabola equation.
We have:
\[{{\left( 2\cos \theta \right)}^{2}}+1=\sin \theta \]
\[4{{\cos }^{2}}\theta +1=\sin \theta \]
Now, applying the trigonometry identity \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \], we will have:
\[4\left( 1-{{\sin }^{2}}\theta \right)+1=\sin \theta \]
\[4{{\sin }^{2}}\theta +\sin \theta -5=0\]
\[4{{\sin }^{2}}\theta +5\sin \theta -4\sin \theta -5=0\]
Factoring the above equation, we will have:
\[\left( 4\sin \theta +5 \right)\left( \sin \theta -1 \right)=0\]
\[\sin \theta =-\dfrac{5}{4}\] (or) \[\sin \theta =1\]
So, the eccentricity is \[\theta ={{\sin }^{-1}}=\dfrac{\pi }{2}\].
As \[\sin \theta =-\dfrac{5}{4}\] is not possible, as it does not lie within the range of the function.
The range of \[\sin \theta \] and \[\cos \theta \] functions is \[[-1,1]\] only, so keep this in mind while solving trigonometric equations.
So, the eccentricity is \[\dfrac{\pi }{2}\].
Hence, option c is the correct answer.
Note: The range of \[\sin \theta \] and \[\cos \theta \] functions is \[[-1,1]\] only, so keep this in mind while solving trigonometric equations.
Last updated date: 24th Sep 2023
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