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HYPERBOLA (EAMCET) EXERCISE

Authored by VASU PAGOTI

Mathematics

12th Grade

Used 5+ times

HYPERBOLA (EAMCET) EXERCISE
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100 questions

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1.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

If the latus rectum of an hyperbola be 8 and eccentricity be 3/√5, then the equation of the hyperbola is

4x2−5y2=1004x^2 - 5y^2 = 100

5x2−4y2=1005x^2 - 4y^2 = 100

4x2+5y2=1004x^2 + 5y^2 = 100

5x2+4y2=1005x^2 + 4y^2 = 100

2.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

The length of the transverse axis of a hyperbola is 7 and it passes through the point (5, -2). The equation of the hyperbola is

449x2−19651y2=1\frac{4}{49} x^2 - \frac{196}{51} y^2 = 1

49/4x2−51/196y2=149/4 x^2 - 51/196 y^2 = 1

449x2−51196y2=1\frac{4}{49} x^2 - \frac{51}{196} y^2 = 1

None of these

3.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

If (0, ± 4) and (0, ± 2) be the foci and vertices of a hyperbola, then its equation is

x2/4−y2/12=1x^2/4 - y^2/12 = 1

x2/12−y2/4=1x^2/12 - y^2/4 = 1

y2/4−x2/12=1y^2/4 - x^2/12 = 1

y2/12−x2/4=1y^2/12 - x^2/4 = 1

4.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

The equation of the hyperbola whose directrix is x + 2y = 1, focus (2, 1) and eccentricity 2 will be

x2−16xy−11y2−12x+6y+21=0x^2 - 16xy - 11y^2 - 12x + 6y + 21 = 0

3x2+16xy+15y2−4x−14y−1=03x^2 + 16xy + 15y^2 - 4x - 14y - 1 = 0

x2+16xy+11y2−12x−6y+21=0x^2 + 16xy + 11y^2 - 12x - 6y + 21 = 0

None of these

5.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

The equation of a hyperbola, whose foci are (5, 0) and (-5, 0) and the length of whose conjugate axis is 8, is

9x2−16y2=1449x^2 - 16y^2 = 144

16x2−9y2=14416x^2 - 9y^2 = 144

9x2−16y2=129x^2 - 16y^2 = 12

16x2−9y2=1216x^2 - 9y^2 = 12

6.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

The equation of the hyperbola referred to its axes as axes of coordinate and whose distance between the foci is 16 and eccentricity is √2, is

x2−y2=16x^2 - y^2 = 16

x2−y2=32x^2 - y^2 = 32

x2−2y2=16x^2 - 2y^2 = 16

y2−x2=16y^2 - x^2 = 16

7.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

The equation of the hyperbola referred to the axis as axes of co-ordinate and whose distance between the foci is 16 and eccentricity is √2, is

x2−y2=16x^2 - y^2 = 16

x2−y2=32x^2 - y^2 = 32

x2−2y2=16x^2 - 2y^2 = 16

y2−x2=16y^2 - x^2 = 16

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