The ends of latus rectum of parabola x 2-8 y -0 areA- -4-2 and 4-2B- -4-2 and 4-2C- 4-2 and -4-2D- 4-2 and -4-2

The correct option is A ( 4, 2) and (4, 2) x^2 = 8y ⇒ a = 2 . So , focus = (0, 2) Ends of latus rectum = (4, 2) , ( 4, 2) . Trick: Since the ends of latus ...

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Standard XI

Mathematics

Terminologies Related to a Parabola

The ends of l...

Question

The ends of latus rectum of parabola $x^{2} + 8 y = 0$ are

(-4, -2) and (4, 2)

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(4, -2) and (-4, 2)

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(-4, -2) and (4, -2)

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(4, 2) and (-4, 2)

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Solution

The correct option is C
(-4, -2) and (4, -2)

$x^{2} = - 8 y \Rightarrow a = - 2 .$ So , focus = (0,-2)

Ends of latus rectum = (4,-2) , (-4,-2) .

Trick: Since the ends of latus rectum lie on parabola , so only points (-4,-2) and (4,-2) satsify the parabola.

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The ends of latus rectum of parabola x2+8y=0 are

The correct option is C (-4, -2) and (4, -2) x2=−8y⇒a=−2. So , focus = (0,-2) Ends of latus rectum = (4,-2) , (-4,-2) . Trick: Since the ends of latus rectum lie on parabola , so only points (-4,-2) and (4,-2) satsify the parabola.

The ends of latus rectum of parabola $x^{2} + 8 y = 0$ are

The correct option is C
(-4, -2) and (4, -2)

$x^{2} = - 8 y \Rightarrow a = - 2 .$ So , focus = (0,-2)

Ends of latus rectum = (4,-2) , (-4,-2) .

Trick: Since the ends of latus rectum lie on parabola , so only points (-4,-2) and (4,-2) satsify the parabola.