Question

The eccentric angle of point of intersection of the ellipse $$x^2+4y^2=4$$ and the parabola $$x^2+1=y$$ is;
(A) $$0$$
(B) $${\cos }^{-1}(-{\displaystyle \frac{2}{3}})$$
(C) $${\displaystyle \frac{\pi }{2}}$$
(D) $${\displaystyle \frac{5\pi }{4}}$$

Answer 1

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 $$x^2+4y^2=4$$, can be rewritten as:
$${\displaystyle \frac{x^2}{4}}+{\displaystyle \frac{y^2}{1}}=1$$
As we know, for any given eccentricity ‘$$\theta$$’, a variable point on ellipse $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$ can be considered as $$(a\cos \theta ,b\sin \theta )$$, now for $${\displaystyle \frac{x^2}{4}}+{\displaystyle \frac{y^2}{1}}=1$$, the variable point on the ellipse will be $$(2\cos \theta ,\sin \theta )$$.
As the ellipse intersects the parabola $$x^2+1=y$$.
The point $$(2\cos \theta ,\sin \theta )$$ 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:
$${(2\cos \theta )}^2+1=\sin \theta$$
$$4{\cos }^2\theta +1=\sin \theta$$
Now, applying the trigonometry identity $${\cos }^2\theta =1-{\sin }^2\theta$$, we will have:
$$4(1-{\sin }^2\theta )+1=\sin \theta$$
$$4{\sin }^2\theta +\sin \theta -5=0$$
$$4{\sin }^2\theta +5\sin \theta -4\sin \theta -5=0$$
Factoring the above equation, we will have:
$$(4\sin \theta +5)(\sin \theta -1)=0$$
$$\sin \theta =-{\displaystyle \frac{5}{4}}$$ (or) $$\sin \theta =1$$
So, the eccentricity is $$\theta ={\sin }^{-1}={\displaystyle \frac{\pi }{2}}$$.
As $$\sin \theta =-{\displaystyle \frac{5}{4}}$$ 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 $${\displaystyle \frac{\pi }{2}}$$.
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.