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}{2}

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\frac{π}{6}

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\frac{π}{3}

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Solution

The correct option is B $\frac{π}{2}$
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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The eccentric angle of point of intersection of the ellipse x2+4y2=4 and the parabola x2+1=y is

The correct option is B π2Since ellipse and parabola intersect at a point ∴y−1=4−4y2 ⇒y−1=4(1−y)(y+1) y=1 or y=−54 Since, y=−54, do not satisfies the equation of parabola ∴y=1 Now, using parametric equations y=0+1⋅sinθ ⇒sinθ=1 ⇒θ=π2

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