# 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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