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Exercise 7.2 · Q38

Q.Find the equation of the ellipse in standard form if the latus rectum has length 6 and foci are (±2,0)(\pm 2, 0).

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Foci (±2,0)(\pm2,0) give ae=2ae=2.

Latus rectum =2b2a=6⇒b2=3a=\dfrac{2b^2}{a}=6 \Rightarrow b^2=3a.

Also b2=a2(1−e2)=a2−a2e2=a2−(ae)2=a2−4b^2=a^2(1-e^2)=a^2-a^2e^2=a^2-(ae)^2=a^2-4.

So 3a=a2−4⇒a2−3a−4=0⇒(a−4)(a+1)=0⇒a=43a=a^2-4 \Rightarrow a^2-3a-4=0 \Rightarrow (a-4)(a+1)=0 \Rightarrow a=4 (rejecting the negative root). …

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