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જેની ઉત્કેન્દ્રતા $e = \frac{1}{2}$ તથા એક નિયામિકા $x=4$ હોય તેવા ઊગમબિંદુ કેન્દ્ર હોય તેવા ઉપવલયનું સમીકરણ મેળવો.
$4{x^2} + 3{y^2} = 1$
$3{x^2} + 4{y^2} = 12$
$4{x^2} + 3{y^2} = 12$
$3{x^2} + 4{y^2} = 1$
Solution
(b) Since directrix is parallel to $y$ – axis,
hence axes of the ellipse are parallel to $x$ – axis.
Let the equation of the ellipse be $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, $(a > b)$
${e^2} = 1 – \frac{{{b^2}}}{{{a^2}}}$
$\Rightarrow \frac{{{b^2}}}{{{a^2}}} = 1 – {e^2} = 1 – \frac{1}{4}$
$\Rightarrow \frac{{{b^2}}}{{{a^2}}} = \frac{3}{4}$.
Also, one of the directrices is $x = 4$
$ \Rightarrow $$\frac{a}{e} = 4$
$\Rightarrow a = 4e = 4.\frac{1}{2} = 2$;
${b^2} = \frac{3}{4}{a^2} = \frac{3}{4}.4 = 3$
$\therefore $ Required ellipse is $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1$
or $3{x^2} + 4{y^2} = 12$.
