If the lengths of the latus rectums of the ellipse E1:4(x+y−1)2+2(x−y+3)2=8 and the ellipse E2:x2k+y2k2=1(0<k<1) are equal, then which of the following is/are true?

Given ellipse 8(x+y−12)2+4(x−y+32)2=8 (x+y−12)2(1)2+(x−y+32)(2)22=1 ⇒2(12)=2(k2k) ⇒k3/2=2−1/2 ⇒k=2−1/3 (ii) Area of the ellipse is πab=πk.k=π(2−1/3)3/2=π2

If the lengths of the latus rectums of the ellipse E1:4(x+y−1)2+2(x−y+3)2=8 and the ellipse E2:x2k+y2k2=1(0<k<1) are equal, then which of the following is/are true?

Given ellipse 8(x+y−12)2+4(x−y+32)2=8 (x+y−12)2(1)2+(x−y+32)(2)22=1 ⇒2(12)=2(k2k) ⇒k3/2=2−1/2 ⇒k=2−1/3 (ii) Area of the ellipse is πab=πk.k=π(2−1/3)3/2=π2

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If the lengths of the latus rectums of the ellipse $E_{1} : 4 {(x + y - 1)}^{2} + 2 {(x - y + 3)}^{2} = 8$ and the ellipse $E_{2} : \frac{x^{2}}{k} + \frac{y^{2}}{k^{2}} = 1 (0 < k < 1)$ are equal, then which of the following is/are true?

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Area of the ellipse $E_{2}$ is $\frac{π}{\sqrt{2}}$

Area of the director circle of $E_{2}$ is $π (2^{\frac{- 1}{3}} + 2^{\frac{- 2}{3}})$ sq.units

Area of the ellipse $E_{2}$ is $\sqrt{2} π$ sq.units

Area of the ellipse $E_{1}$ is $\sqrt{2} π$ sq.units

answer is A, B, D.

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Detailed Solution

Given ellipse $8 {(\frac{x + y - 1}{\sqrt{2}})}^{2} + 4 {(\frac{x - y + 3}{\sqrt{2}})}^{2} = 8$
$\frac{{(\frac{x + y - 1}{\sqrt{2}})}^{2}}{{(1)}^{2}} + {\frac{(\frac{x - y + 3}{\sqrt{2}})}{{(\sqrt{2})}^{2}}}^{2} = 1$
$\Rightarrow 2 (\frac{1}{\sqrt{2}}) = 2 (\frac{k^{2}}{\sqrt{k}}) \Rightarrow k^{3 / 2} = 2^{- 1 / 2} \Rightarrow k = 2^{- 1 / 3}$
(ii) Area of the ellipse is $π a b = π \sqrt{k} . k = π {(2^{- 1 / 3})}^{3 / 2} = \frac{π}{\sqrt{2}}$