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

181.

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


182.

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


Advertisement

183.

If two circles 2x2 + 2y2 - 3x + 6y + k = 0 and x2 + y2 - 4x + 10y + 16 = 0 cut orthagobally then, value of k is

  • 41

  • 14

  • 4

  • 1


C.

4

We have given two circles are

2x2 + 2y2 - 3x + 6y + k = 0

   x2 + y2 - 32x + 3y + k2 = 0         ...(i)and x2 + y2 - 4x + 10y + 16 = 0      ...(ii)

Since, general equation of circle is

x2 + y2 + 2gx + 2fy + c = 0                               ...(iii)

Therefore, comparing Eqs. (i) and (ii) with Eq. (iii), we get

       g1 = - 34, f1 = 32, c1 = k2and g2 = - 2, f2 = 5, c2 = 16Since, both the circles cut orthagonally. 2g1g2 + f1f2 = c1 + c2    232 + 152 = k2 + 16                    18 = k2 + 16                    k2 = 2                     k = 4


Advertisement
184.

The locus of z satisfying the inequality z +2i2z +i < 1, where z = x + iy, is

  • x2 + y2 < 1

  • x2 - y2 < 1

  • x2 + y2 > 1

  • 2x2 + 3y2 < 1


Advertisement
185.

The area (in square unit) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is

  • 4π

  • 3π

  • 2π

  • π


186.

The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the x and y -axes respectively are

  • x2 + y2 ± 4x ± 8y = 0

  • x2 + y2 ± 2x ± 4y = 0

  • x2 + y2 ± 8x ± 16y = 0

  • x2 + y2 ± x ± y = 0


187.

The point (3 -4) lies on both the circles x2 + y2 - 2x + 8y + 13 = 0 and x2 + y2 - 4x + 6y + 11 = 0. Then, the angle between the circles is

  • 60°

  • tan-112

  • tan-135

  • 135°


188.

The equation of the circle which passes through the origin and cuts orthogonally each of the circles x+ y2 - 6x + 8 = 0 and x2 + y2 - 2x - 2y = 7 is

  • 3x2 + 3y2 - 8x - 13y = 0

  • 3x2 + 3y2 - 8x + 29y = 0

  • 3x2 + 3y2 + 8x + 29y = 0

  • 3x2 + 3y2 - 8x - 29y = 0


Advertisement
189.

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

  • 0

  • 1

  • 2

  • 3


190.

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


Advertisement