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Exercise 7.2 · Q40

Q.Find the equation of the ellipse in standard form if the dist. between its directrix is 10 and which passes through (−5,2)(-\sqrt5, 2).

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2ae=10⇒a=5e\dfrac{2a}{e}=10 \Rightarrow a=5e.

The point (−5,2)(-\sqrt5,2) lies on x2a2+y2a2(1−e2)=1\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2(1-e^2)}=1: 5a2+4a2(1−e2)=1\dfrac{5}{a^2}+\dfrac{4}{a^2(1-e^2)}=1.

Substituting a=5ea=5e (so a2=25e2a^2=25e^2) and solving: 525e2+425e2(1−e2)=1\dfrac{5}{25e^2}+\dfrac{4}{25e^2(1-e^2)}=1, which reduces to e2=1525=35e^2=\dfrac{15}{25}=\dfrac35, so e=155e=\dfrac{\sqrt{15}}{5}. …

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