The equation of the common tangents to the two hyperbolas x2

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 Multiple Choice QuestionsMultiple Choice Questions

181.

If x2a2 + y2b2 = 1 a > b and x2 - y2 = c2 cut at right angles, then

  • a2 + b2 = 2c2

  • b2 - a2 = 2c2

  • a2 - b2 = 2c2

  • a2b2 = 2c2


182.

The equation of the conic with focus at (1, - 1) directrix along x - y + 1 = 0 and with eccentricity 2, is

  • x2 - y2 = 1

  • xy = 1

  • 2xy - 4x + 4y + 1 = 0

  • 2xy + 4x - 4y - 1 = 0


183.

The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 - 6x - 8y = 24 is

  • 0

  • 1

  • 3

  • 4


184.

The locus of the mid-points of the focal chord of the parabola y2 = 4ax is

  • y2 = a(x - a)

  • y2 = 2a(x - a)

  • y2 = 4a(x - a)

  • None of these


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185.

A rod of length l slides with its ends on two perpendicular lines. Then, the locus of its mid point is

  • x2 + y2 = l24

  • x2 + y2 = l22

  • x2 - y2 = l24

  • None of these


186.

The line joining (5, 0) to (10cosθ, 10sinθ) is divided internally in the ratio 2 : 3 at P. If 0 varies, then the locus of P is

  • a straight line

  • a pair of straight lines

  • a circle

  • None of the above


187.

If 2x + y + k = 0 is a normal to the parabola y2 = - 8x, then the value of k, is

  • 8

  • 16

  • 24

  • 32


188.

If the equation of an ellipse is 3x2 + 2y2 + 6x - 8y + 5 = 0, then which of the following are true?

  • e = 13

  • centre is (- 1, 2)

  • foci are (- 1, 1) are (- 1, 3)

  • All of the above


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189.

The equation of the common tangents to the two hyperbolas x2a2 - y2b2 = 1 and y2a2 - x2b2 = 1, are

  • y = ± x ± b2 - a2

  • y = ± x ± a2 - b2

  • y = ± x ± a2 + b2

  • y = ± x ± a2 - b2


B.

y = ± x ± a2 - b2

We have the hyperbolas

       x2a2 - y2b2 = 1           ...(i)

and y2a2 - x2b2 = 1            ...(ii)

Any tangent to the hyperbola Eq. (i),

          y = mx + c

where c = ± a2m2 - b2        ...(iii)

But this tangent touches the parabola Eq. (ii), also

 mx + c2a2 - x2b2 = 1 b2m2x2 + c2 + 2cmx - a2x2 = a2b2 b2m2 - a2x2 + 2mcxb2 + b2c2 - a2b2 = 0  b2m2 - a2x2 + 2mcb2x + b2c2 - a2 = 0For the tangency, it should have equal roots2mcb22 = 4b2m2 - a2 . b2c2 - a2

 4m2c2b4 = 4b2b2m2c2 - b2m2a2 - a2c2 + a4  m2c2b2 = b2m2c2 - b2m2a2 - a2c2 + a4      a2c2 = a4 -  b2m2a2          c2 = a2 - b2m2   a2m2 - b2 = a2 - b2m2        using Eq. (iii) a2 + b2m2 =a2 + b2                m2 = 1                 m = ± 1

Hence, the equation of common tangent are

                    y = ± x + ± a2 - b2

 


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190.

The equation of sphere concentric with the sphere x2 + y2 + z2 - 4x - 6y - 8z - 5 = 0 and which passes through the origin, is

  • x2 + y2 + z2 - 4x - 6y - 8z = 0

  • x2 + y2 + z2 - 6y - 8z = 0

  • x2 + y2 + z2 = 0

  • x2 + y2 + z2 - 4x - 6y - 8z - 6 = 0


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