Question

Which lines are the directrices of the ellipse?

x = −4.25 and x = 8.25
x = −3.25 and x = 9.25
y = −4.25 and y = 8.25
y = −3.25 and y = 9.25

Answer 1

$\textbf{To find:}$

$\textsf{Equation of directrices of the given ellipse}$

$\textbf{Solution:}$

$\textsf{From the figure, 2a=10, 2b=6}$

$\mathsf{and\;centre\;is\;(3,2)}$

$\implies\mathsf{a=5\;and\;b=3}$

$\textsf{The equation of ellipse is }$

$\mathsf{\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1}$

$\mathsf{\dfrac{(x-3)^2}{5^2}+\dfrac{(y-2)^2}{3^2}=1}$

$\mathsf{\dfrac{(x-3)^2}{25}+\dfrac{(y-2)^2}{9}=1}$

$\mathsf{Eccentricity,\;e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{25-9}{25}}=\sqrt{\dfrac{16}{25}}}$

$\implies\mathsf{e=\dfrac{4}{5}}$

$\mathsf{\dfrac{a}{e}=\dfrac{5}{\dfrac{4}{5}}=\dfrac{25}{4}}$

$\mathsf{Equation\;of\;directrices\;are}$

$\mathsf{x-3=\pm\,\dfrac{25}{4}}$

$\mathsf{x-3=\pm\,6.25}$

$\mathsf{x=3\,\pm\,6.25}$

$\mathsf{x=3+6.25\;and\;x=3-6.25}$

$\implies\boxed{\mathsf{x=9.25\;and\;x=-3.25}}$

Find more:

Eccentricity of the ellipse whose latus rectum is equal to the distance between two focus points, is

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