If the lines 2x + 3y +12 = 0, x - yy + k = 0 are conjugate with&n

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 Multiple Choice QuestionsMultiple Choice Questions

531.

The equation of the radical axis of the pair of circles  7x2 + 7y2 - 7x + 14y + 18 = 0  and 4x2 + 4y2 - 7x + 8y + 20 = 0 is

  • x - 2y - 5 = 0

  • 2x - y + 5 = 0

  • 21x - 68 = 0

  • 23x - 68 = 0


532.

x2 + y2 - 8x + 40 = 05x2 + 5y2 -25x + 80 = 0x2 + y2 - 8x + 16y + 160 = 0From the point P are equal, then P = ?

  • 8, 152

  • - 8, 152

  • 8, - 152

  • - 8, - 152


533.

The equation of the circle concentric with the circle x2 + y2 - 6x + 12y + 15 = 0 and of double its area is

  • x2 + y2 - 6x +12y - 15 = 0

  • x2 + y2 - 6x +12y - 30 = 0

  • x2 + y2 - 6x +12y - 25 = 0

  • x2 + y2 - 6x +12y - 20 = 0


534.

If the circle x2 + y2 + 2x + 3y + 1 = 0 cuts another circle x2 + y+ 4x + 3y + 2 = 0 in A and B, then the equation of the circle with AB as a diameter is

  • x2 + y2  + x + 3y + 1 = 0

  • 2x2 + 2y2  + 2x + 6y + 1 = 0

  • x2 + y2  + x + 6y + 1 = 0

  • 2x2 + 2y2  + x + 3y + 1 = 0


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535.

The equation of the hyperbola which passes through the point (2, 3) and has the asymptotes 4x + 3y - 7 = 0 and x - 2y - 1 = 0 is

  • 4x2 + 5xy - 6y2 - 11x + 11y + 50 = 0

  • 4x2 + 5xy - 6y2 - 11x + 11y - 43 = 0

  • 4x2 - 5xy - 6y2 - 11x + 11y + 57 = 0

  • x2 - 5xy - y2 - 11x + 11y - 43 = 0


536.

The product of the perpendicular distances from any point on the hyperbola x2a2 - y2b2 = 1 to its asymtotes is

  • a2b2a2 - b2

  • a2b2a2 + b2

  • a2 + b2a2b2

  • a2 - b2a2b2


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537.

If the lines 2x + 3y +12 = 0, x - yy + k = 0 are conjugate with respect to the parabola y2 = 8x, then k is equal to

  • 10

  • 72

  • - 12

  • - 2


C.

- 12

Given, conjugate lines are 2x +3y + 12 = 0    ...iand x - y + k = 0    ...ii

We know, that two lines are said to be conjugate with respect to a curve, if each passes through the pole of the polar of that curve. Let (x1, y1) be the pole of parabola

y2 = 8x

Its polar is yy1 = 4x + x1 4x - y1y + 4x1 = 0 2x - y12y + 2x1 = 0      ...iiiOn comparing Eqs. (i) and (iii), we get- y12 = 3  y1 = - 6 and 2x1 = 12  x1 = 6 Pole x1, y1 = 6, - 6Eq. (ii) also passes through pole (6, - 6) 6 - - 6 + k = 0                      k = - 12


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538.

Find the equation to the parabola, whose axis parallel to they-axis and which passes through the points (0, 4), (1, 9) and (4, 5) is

  • y = - x+ x + 4

  • y = - x+ x + 1

  • y = - 1912x2 + 7912x + 4

  • y = - 1912x2 + 8912x + 4


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539.

If the line y = 2x + c is a tangent to the circle x2 + y2 = 5, then a value of

  • 2

  • 3

  • 4

  • 5


540.

A line segment AM = a moves in the XOY plane such that AM is parallel to the X-axis. If A moves along the circle x2 + y= a2, then the locus of M is

  • x2 + y= 4a2

  • x2 + y= 2ax

  • x2 + y= 2ay

  • x2 + y= 2ax + 2ay


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