The length of the parabola y2 = 12x cut off by the latusrectum is

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 Multiple Choice QuestionsMultiple Choice Questions

391.

The number of normals drawn to the parabola y2 = 4x from the point (1, 0) is

  • 0

  • 1

  • 2

  • 3


392.

If the circle x2 + y2 = a intersects the hyperbola xy = cin four points (xi, yi), for i = 1, 2, 3 and 4, then y1 + y2 + y3 + y4 equals

  • 0

  • c

  • a

  • c4


393.

The mid point of the chord 4x - 3y = 5 of the hyperbola 2x- 3y2 = 12 is

  • 0, - 53

  • (2, 1)

  • 54, 0

  • 114, 2


394.

The radius of the sphere x2 + y2 + z2 = 12x + 4y + 3z is

  • 13/2

  • 13

  • 26

  • 52


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395.

The centre and radius of the sphere x2 + y2 + z2 + 3x - 4z + 1 = 0 are

  • - 32, 0, - 2; 212

  • 32, 0, 2; 21

  • - 32, 0, 2; 212

  • - 32, 0, 2; 212


396.

Let A and B are two fixed points in a plane, then locus of another point Con the same plane such that CA + CB = constant, (> AB) is

  • circle

  • ellipse

  • parabola

  • hyperbola


397.

The directrix of the parabola y2 + 4x + 3 = 0 is

  • x - 43 = 0

  • x + 14 = 0

  • x - 34 = 0

  • x - 14 = 0


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398.

The length of the parabola y2 = 12x cut off by the latusrectum is

  • 62 + log1 + 2

  • 32 + log1 + 2

  • 62 - log1 + 2

  • 32 - log1 + 2


A.

62 + log1 + 2

Given equation of parabola is

         y2 = 12x         ...(i)

and equation of latusrectum is

           x = 3             ...(ii)

From Eqs. (i) and (ii), we get

         y2 = 36

        y = ± 6

 Coordinates of end points of a latusrectum are (3, 6) and (3, - 6).

 Required length = 2031 + dydx2dx= 2031 + 6y2= 20312x + 3612xdx= 203x + 3x2 + 3xdx= 2x2 + 3x + 32logx + 32 + x2 + 3x03= 232 + 32log92 + 32 - 32log32= 232 + 3log2 + 1= 62 + log1 + 2


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399.

Area enclosed by the curve π4x - 22 + y2 = 8 is

  • π sq unit

  • 2 sq unit

  • 3π sq unit

  • 4 sq unit


400.

The equation of a directrix of the ellipse x216 + y225 = 1 is :

  • 3y = 5

  • y = 5

  • 3y = 25

  • y = 3


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