(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the foci are $(\pm 2, 0)$,the ellipse is horizontal with foci at

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$\frac{x^2}{16} + \frac{y^2}{12} = 1$

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the foci are $(\pm 2, 0)$,the ellipse is horizontal with foci at $(\pm ae, 0)$.Thus,$ae = 2$. Given $e = 1/2$,we have $a(1/2) = 2$,which implies $a = 4$.Using the relation $b^2 = a^2(1 - e^2)$:$b^2 = 16(1 - (1/2)^2) = 16(1 - 1/4) = 16(3/4) = 12$.Substituting $a^2 = 16$ and $b^2 = 12$ into the standard equation,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.

Class 11 Mathematics 10-2. Parabola, Ellipse, Hyperbola Ellipse

Easy

Find the equation of the ellipse whose foci are $(\pm 2, 0)$ and eccentricity is $1/2$.

A
$\frac{x^2}{16} + \frac{2y^2}{9} = 1$
B
$\frac{x^2}{12} + \frac{y^2}{10} = 1$
C
$\frac{x^2}{16} + \frac{y^2}{12} = 1$
D
$\frac{3x^2}{12} + \frac{y^2}{14} = 1$

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10-2. Parabola, Ellipse, Hyperbola

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Ellipse

Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ be a given ellipse whose latus rectum length is $10$. If its eccentricity $e$ is the maximum value of the function $\phi(t) = \frac{5}{12} + t - t^{2}$,then $a^{2} + b^{2}$ is equal to:

DifficultJEE MAIN 2020

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Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^{2}-b^{2}}, 0)$ and $(-\sqrt{a^{2}-b^{2}}, 0)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^{2}$.

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The equations of the tangents to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ making an inclination of $30^\circ$ with the major axis are

DifficultMHT CET 2026

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The area of the region bounded by the curve $9x^2 + 4y^2 - 36 = 0$ is (in $\pi$)

Medium

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If $\tan \theta_1 \times \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining $2$ points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at

MediumAP EAMCET 2021

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Find the equation of the ellipse whose foci are $(\pm 2, 0)$ and eccentricity is $1/2$.

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Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ be a given ellipse whose latus rectum length is $10$. If its eccentricity $e$ is the maximum value of the function $\phi(t) = \frac{5}{12} + t - t^{2}$,then $a^{2} + b^{2}$ is equal to:

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