Courses
Courses for Kids
Free study material
Free LIVE classes
More

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) $\dfrac{5\pi }{4}$

Last updated date: 18th Mar 2023
Total views: 305.1k
Views today: 7.84k
Verified
305.1k+ views
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.

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.