The eccentric angle of point of intersection of the ellipse $x^{2} + 4 y^{2} = 4$ and the parabola $x^{2} + 1 = y$ is

\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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\frac{π}{6}

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Solution

The correct option is C $\frac{π}{2}$
Since ellipse and parabola intersect at a point
$∴ y - 1 = 4 - 4 y^{2}$
$\Rightarrow y - 1 = 4 (1 - y) (y + 1)$
$y = 1$ or $y = - \frac{5}{4}$
Since, $y = - \frac{5}{4}$ , do not satisfies the equation of parabola
$∴ y = 1$
Now, using parametric equations $y = 0 + 1 \cdot sin θ$
$\Rightarrow sin θ = 1$
$\Rightarrow θ = \frac{π}{2}$

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Similar questions

x^{2} + 4 y^{2} = 4

x^{2} + 1 = y

A. If the tangent to the ellipse $x^{2} + 4 y^{2} = 16$ at the point $P (ϕ)$ is a normal to the circle $x^{2} + y^{2} - 8 x - 4 y = 0$ , then $\frac{ϕ}{2}$ may be	1. $0$
B. The eccentric angle(s) of a point on the ellipse $x^{2} + 3 y^{2} = 6$ at a distance $2$ units from the centre of the ellipse is/are	2. ${cos}^{- 1} (- \frac{2}{3})$
C. The eccentric angle of intersection of the ellipse $x^{2} + 4 y^{2} = 4$ and the parabola $x^{2} + 1 = y$ is	3. $\frac{π}{4}$
D. If the normal at the point $P (θ)$ to the ellipse $\frac{x^{2}}{14} + \frac{y^{2}}{5} = 1$ intersects it again at the point $Q (2 θ)$ , then $θ$ is	4. $\frac{5 π}{4}$