If the normal to the parabola y2 = 4x at P(1, 2) meets the parabo

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

261.

If the line x + 2by + 7 = 0 is a diameter of the circle x2 + y2 - 6x + zy = 0, then b is equal to

  • - 5

  • - 3

  • 2

  • 5


262.

If the line y = 2x+ c is tangent to the parabola y2 = 4x, then c is equal to

  • - 12

  • 12

  • 13

  • 14


263.

The length of the latus rectum of the ellipse 9x2 + 4y2 = 1 is

  • 32

  • 49

  • 83

  • 89


264.

The number of circles that touch all the straight lines x + y = 4,x - y = - 2 and y = 2 is

  • 1

  • 2

  • 3

  • 4


Advertisement
265.

The equation of the normal to the circle x2 + y2 + 6x + 4y - 3 = 0 at (1, - 2) is

  • y + 1 = 0

  • y + 2 = 0

  • y + 3 = 0

  • y - 2 = 0


266.

The limiting points of the co-axial system containing the two circles x2 + y2 + 2x - 2y + 2 = 0 and 25(x2 + y) - 10x - 80y + 65 = 0 are

  • (1, - 1), (- 3, - 40)

  • 1, - 1, - 15, 85

  • - 1, 1, 15, 85

  • - 15, - 85


267.

The radical axis of circles x2 + y2 + 5x + 4y - 5 = 0 and x2 + y2 - 3x + 5y - 6 = 0 is

  • 8y - x + 1 = 0

  • 8x - y+ 1 = 0

  • 8x - 8y + 1 = 0

  • y - 8x + 1 = 0


268.

The length of latusrectum of parabola y2 + 8x - 2y + 17 = 0 is

  • 2

  • 4

  • 8

  • 16


Advertisement
Advertisement

269.

If the normal to the parabola y2 = 4x at P(1, 2) meets the parabola again in Q, then coordinates of Q are

  • (- 6, 9)

  • (9, - 6)

  • (- 9, - 6)

  • (- 6, - 9)


B.

(9, - 6)

The given equation of the parabola is          y2 = 4xEquation of tangent at (1, 2) is y . 2 = 2x + 1      y = x + 1Slope of tangent = 1 slope of the normal = - 1Equation of the normal passing through (1, 2) is      y - 2 = - 1x - 1 y - 2 = - x + 1  x +y = 3This equation again meet at Q of the parabola, then     y2 = 43 - y y2 = 12 - 4y  y2 + 4y - 12 = 0 y - 2y + 6 = 0  y = 2, - 6If y = - 6, then x = 3 - - 6 = 9 The coordinates of point Q are (9, - 6).


Advertisement
270.

The eccentricity of ellipse x216 + y29 = 1 is

  • 716

  • 54

  • 74

  • 72


Advertisement