If the length of the latus rectum of the ellipse x2 + 4y2 + 2x + 8y ? = 0 is 4, and l is the length of its major axis, then + l is equal to:

Question

If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to:

Smart Achievers · Accepted Answer

(75)

$We have x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$

$\Rightarrow \frac{{(x + 1)}^{2}}{λ + 5} + \frac{{(y + 1)}^{2}}{\frac{λ + 5}{4}} = 1$

$∵$ $\frac{2 b^{2}}{a} = 4$

$\frac{2 (λ + 5)}{4} = 4 \sqrt{λ + 5}$

$On solving \Rightarrow λ = 59$

$λ = - 5$

$l = 2 a = 2 \sqrt{λ + 5} = 2 \sqrt{65} = 16$

$\Rightarrow λ + l = 59 + 16 = 75$

Solution

Q.
If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to:

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Solution

Q. If the length of the latus rectum of the ellipse x2+4y2+2x+8y-λ=0 is 4, and l is the length of its major axis, then λ+l is equal to:

Ans. (75) We have x2+4y2+2x+8y-λ=0 ⇒(x+1)2λ+5+(y+1)2λ+54=1 ∵ 2b2a=4 2(λ+5)4=4λ+5 On solving ⇒λ=59 λ=-5 l=2a=2λ+5=265=16 ⇒λ+l=59+16=75

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Q.
If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to: