The distance between the foci of a hyperbola is 16 and its eccent

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

421.

If the line 3x - 2y + p = 0 is normal to the circle x2 + y2 = 2x - 4y - 1, then p will be

  • - 5

  • 7

  • - 7

  • 5


422.

If the two circles x2 + y2 = r2 and x2 + y2 - 10x + 16 = 0 intersect at two real points, then

  • 1 < r < 7

  • 3 < r < 10

  • 2 < r < 9

  • 2 < r < 8


423.

The equation of the common tangent to the parabolas y2 = 2x and x2 = 16y will be

  • x + y + 2 = 0

  • x - 3y + 1 = 0

  • x + 2y - 2 = 0

  • x + 2y + 2 = 0


424.

The equation of the tangent to the parabola y2 = 8x, which is parallel to the line 2x - y + 7 = 0, will be

  • y = x + 1

  • y = 2x + 1

  • y = 3x + 1

  • y = 4x + 1


Advertisement
425.

The distance of a point on ellipse x26 + y22 = 1 from its centre is 2. The eccentric angle of the point will be

  • π4 or π3

  • π3 or 3π5

  • π4 or 3π4

  • None of these


426.

The distance between the foci of a hyperbola is 16 and its eccentncity is 2. Its equation will be

  • x2 - y2 = 1

  • x2 - y2 = 20

  • x2 - y2 = 4

  • x2 - y2 = 32


427.

Equation of the circle which passes through the origin and cuts intercepts of lengths a and b on axes is

  • x2 + y2 + ax + by = 0

  • x2 + y2 + ax - by = 0

  • x2 + y2 + bx + ay = 0

  • None of the above


428.

If focus ofa parabolais at (3, 3) and its directrix is 3x - 4y = 2, then its latusrectum is

  • 2

  • 3

  • 4

  • 5


Advertisement
429.

If the straight line y = 4x + c is a tangent to the ellipse x28 + y24 = 1, then c will be equal to

  • ± 4

  • ± 6

  • ± 1

  • ± 132


Advertisement

430.

The distance between the foci of a hyperbola is 16 and its eccentricity is - 2. Its equation will be

  • x2 - y2 = 32

  • y2 - x2 = 32

  • x2 - y2 = 16

  • y2 - x2 = 16


A.

x2 - y2 = 32

Given, 2ae = 16             ae = 8 a . 2 = 8    e = 2          a = 82 = 42We know that,b2 = a2e2 - 1     = 322 - 1 = 32 Equation of hyperbola is  x2a2 - y2b2 = 1 x2 - y2 = 32


Advertisement
Advertisement