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10-2. Parabola, Ellipse, Hyperbola
easy
ઉગમબિંદુ આગળ કેન્દ્રવાળા ઉપવલયની ઉત્કેન્દ્રતા $1/2$ છે. જો એક નિયામિકા $x = 4$ હોય તો ઉપવલયનું સમીકરણ :
A
$3x^2 + 4y^2 = 1$
B
$3x^2 + 4y^2 = 12$
C
$4x^2 + 3y^2 =12$
D
$4x^2 + 3y^2 = 1$
Solution
Equation of directrix is $x =4$ which is parallel to $y$-axis so major axis of the ellipse is $x$-axis and given center of ellipse is origin. Let equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a > b)$
Given,eccentricity e $=\frac{1}{2}$, we know that distance of directrix from center is $\frac{ a }{ e }=$ $4 \Rightarrow \quad a=2$, from eccentricity definition we get value of $b$ as $\sqrt{3}$,now equation becomes $\frac{x^2}{4}+\frac{y^2}{3}=1 \quad \Rightarrow \quad 3 x^2+4 y^2=12$
Standard 11
Mathematics