Skip to main content
Back to HYPERBOLA

Your Progress

0 / 298 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 61-90 of 298 questions

Preparing practice and saved progress…

61 Easy
61 of 298
Application +4 | -1
Tangents are drawn to the hyperbola 5x2−y2=55 x^2-y^2=5 from the point (0,2)(0,2). Their equations are
View answer and explanation
Correct option: A
y=±3x+2y= \pm 3 x+2
Option A (y=±3x+2y= \pm 3 x+2) is correct because substituting the point (0,2) into the general tangent equation of the hyperbola yields the given equations.
62 Easy
62 of 298
Analyzing +4 | -1
Let P(6,3)P(6,3) be a point on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. If the normal at the point PP intersects the xx-axis at (9,0)(9,0), then the eccentricity of the hyperbola, is
View answer and explanation
Correct option: B
32\sqrt{\frac{3}{2}}
Option B (32\sqrt{\frac{3}{2}}) is correct because by using the equation of the normal to the hyperbola and the given point, we can solve for the eccentricity.
63 Easy
63 of 298
Application +4 | -1
Tangents are drawn to the hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=1 parallel to the straight line 2x−y=12 x-y=1. The points of contact of the tangents on the hyperbola, are
View answer and explanation
Correct option: A
(±922,±12)\left(\pm \frac{9}{2 \sqrt{2}}, \pm \frac{1}{\sqrt{2}}\right)
Option A (±922,±12)\left(\pm \frac{{9}}{{2 \sqrt{{2}}}}, \pm \frac{{1}}{{\sqrt{{2}}}}\right) is correct because it is derived from the formula for points of contact of tangents to a hyperbola.
64 Easy
64 of 298
Application +4 | -1
The equation of the director circle of the hyperbola 9x2−16y2=1449{x^2} - 16{y^2} = 144 is
View answer and explanation
Correct option: A
x2+y2=7{x^2} + {y^2} = 7
Option A (x2+y2=7{x^2} + {y^2} = 7) is correct because the equation of the director circle is x2+y2=a2−b2{x^2} + {y^2} = a^2 - b^2 for a hyperbola.
65 Easy
65 of 298
Application +4 | -1
The equation of the hyperbola whose foci are (6,5),(−4,5)\left( {6,5} \right),\left( { - 4,5} \right) and eccentricity 54\frac{5}{4}, is
View answer and explanation
Correct option: A
(x−1)216−(y−5)29=1\frac{{{{\left( {x - 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 5} \right)}^2}}}{9} = 1
Option A ((x−1)216−(y−5)29=1\frac{{{{\left( {x - 1} \right)}^2}}}{{16}} - \frac{{{{\left( {y - 5} \right)}^2}}}{9} = 1) is correct because it represents the equation of a hyperbola with the given foci and eccentricity, derived by finding the center, a, and b.
66 Easy
66 of 298
Application +4 | -1
For the ellipse 3x2+4y2+6x−8y−5=03{x^2} + 4{y^2} + 6x - 8y - 5 = 0 the eccentricity is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because after completing the squares and transforming the equation into standard form, the eccentricity of the ellipse is calculated using the formula e=1−b2a2e = \sqrt{1 - \frac{b^2}{a^2}}.
67 Easy
67 of 298
Application +4 | -1
Given the points A(0,4)A\left( {0,4} \right) and B(0,−4)B\left( {0, - 4} \right), then the equation of the locus of the point P(x,y)P\left( {x,y} \right) such that, ∣AP−BP∣=6\left| {AP - BP} \right| = 6, is
View answer and explanation
Correct option: D
y29−x27=1\frac{{{y^2}}}{9} - \frac{{{x^2}}}{7} = 1
Option D (y29−x27=1\frac{{{y^2}}}{9} - \frac{{{x^2}}}{7} = 1) is correct because it represents the equation of a hyperbola derived from the given condition ∣AP−BP∣=6\left| {AP - BP} \right| = 6.
68 Easy
68 of 298
Application +4 | -1
If the line x−1=0x-1=0 is a directrix of the hyperbola kx2−y2=6k x^2-y^2=6, then the hyperbola passes through the point
View answer and explanation
Correct option: C
(5,−2)(\sqrt{5},-2)
Option C is correct because by finding the eccentricity and using the directrix equation, we can determine the value of k and then check which point satisfies the hyperbola equation.
69 Easy
69 of 298
Application +4 | -1
The line 93x+12y=23439 \sqrt{3} x+12 y=234 \sqrt{3} is a normal to the hyperbola x281−y236=1\frac{x^2}{81}-\frac{y^2}{36}=1 at the points.
View answer and explanation
Correct option: A
(18,63)(18,6 \sqrt{3})
Option A ((18,63)(18, 6\sqrt{3})) is correct because the point satisfies the equation of the normal to the hyperbola at that point.
70 Easy
70 of 298
Application +4 | -1
The equation of the normal to the hyperbola x216−y29=1\frac{x^2}{16}-\frac{y^2}{9}=1 at the point (8,33)(8,3 \sqrt{3}) is
View answer and explanation
Correct option: D
2x+3y=252 x+\sqrt{3} y=25
Option D (2x+3y=252 x+\sqrt{3} y=25) is correct because substituting the point (8, 3√3) into the equation satisfies the equation, and it represents the normal to the hyperbola at that point.
71 Easy
71 of 298
Application +4 | -1
The product of the lengths of perpendiculars drawn from any point on the hyperbola x2−2y2−2=0{x^2} - 2{y^2} - 2 = 0 to its asymptotes is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because the product of perpendiculars from any point on the hyperbola to its asymptotes is a constant value determined by the hyperbola's parameters.
72 Easy
72 of 298
Application +4 | -1
If lx+my+n=0lx + my + n = 0 is a tangent to the rectangular hyperbola xy=c2,xy = {c^2}, then
View answer and explanation
Correct option: A
l<m<0l < m < 0
Option D (None of these) is correct because the condition for tangency to a rectangular hyperbola leads to a different relationship between l, m, and n.
73 Easy
73 of 298
Application +4 | -1
Focus of hyperbola is (±3,0)\left( { \pm 3,0} \right) and equation of tangent is 2x+y−4=02x + y - 4 = 0, find the equation of hyperbola
View answer and explanation
Correct option: A
4x2−5y2=204{x^2} - 5{y^2} = 20
Option A (4x2−5y2=204{x^2} - 5{y^2} = 20) is correct because it satisfies the given conditions for the foci and the tangent.
74 Easy
74 of 298
Application +4 | -1
If length of the transverse axis of a hyperbola is 8 and its eccentricity is 5/2\sqrt{5} / 2 then the length of a latus rectum of the hyperbola is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the length of the latus rectum is calculated using the formula 2b2a\frac{{2b^2}}{{a}} and the given information.
75 Easy
75 of 298
Application +4 | -1
The eccentricity of the hyperbola in the standard form x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1, passing through (3,0) and (3,2,2)\left( {3,\sqrt 2 ,2} \right) is
View answer and explanation
Correct option: D
133\frac{{\sqrt {13} }}{3}
Option D 133\frac{{\sqrt {13} }}{3} is correct because it follows from substituting the given points into the hyperbola equation and solving for eccentricity using the formula e=1+b2a2e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
76 Easy
76 of 298
Application +4 | -1
The eccentricity of the hyperbola x2−y2=2004{x^2} - {y^2} = 2004 is
View answer and explanation
Correct option: D
2\sqrt 2
Option D (2\sqrt 2) is correct because the given hyperbola is a rectangular hyperbola, and the eccentricity of a rectangular hyperbola is always 2\sqrt{2}.
77 Easy
77 of 298
Application +4 | -1
If the line 2x+6y=22x + \sqrt 6 y = 2 touches the hyperbola x2−2y2=4{x^2} - 2{y^2} = 4, then the point of contact is
View answer and explanation
Correct option: D
(4,−6)(4, - \sqrt {6)}
Option D ((4,−6)(4, - \sqrt {6)}) is correct because the point of contact is found by comparing the tangent equation to the hyperbola with the given line equation.
78 Easy
78 of 298
Application +4 | -1
The angle between the asymptotes of the hyperbola 27x2−9y2=2427{x^2} - 9{y^2} = 24 is
View answer and explanation
Correct option: B
1200{120^0}
Option B (1200{120^0}) is correct because it is calculated using the formula 2 * arctan(b/a) after rewriting the hyperbola equation in standard form.
79 Easy
79 of 298
Application +4 | -1
The eccentricity of the hyperbola 19993(x2−y2)=1,\frac{{\sqrt {1999} }}{3}\left( {{x^2} - {y^2}} \right) = 1, is
View answer and explanation
Correct option: A
2\sqrt 2
Option A (2\sqrt 2) is correct because the given hyperbola is a rectangular hyperbola, and rectangular hyperbolas always have an eccentricity of 2\sqrt 2.
80 Easy
80 of 298
Application +4 | -1
If two perpendicular tangents are drawn from a point (α,β)(\alpha, \beta) to the hyperbola x2−y2=16x^2-y^2=16, then the locus of (α,β)(\alpha, \beta) is
View answer and explanation
Correct option: A
A pair of straight line
The correct answer is A (A pair of straight line) because the locus of the point from which perpendicular tangents are drawn to the hyperbola is a pair of straight lines.
81 Easy
81 of 298
Application +4 | -1
Let the eccentricity of the hyperbola H:x2a2−y2b2=1H:\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 be 52\sqrt {\frac{5}{2}} and length of its latus rectum be 62.6\sqrt 2 . If y=2x+cy = 2x + c is a tangent to the hyperbola H,H, Then the value of c2{c^2} is equal to
View answer and explanation
Correct option: B
20
Option B (20) is correct because the value of c2 is calculated using the eccentricity and latus rectum to find a and b, then applying the tangent condition.
82 Easy
82 of 298
Application +4 | -1
The eccentricity of the hyperbola which passes through (3, 0) and (32,2)(3\sqrt 2 ,2), is
View answer and explanation
Correct option: B
133\frac{{\sqrt {13} }}{3}
Option B 133\frac{{\sqrt {13} }}{3} is correct because it is the eccentricity calculated from the given points on the hyperbola.
83 Easy
83 of 298
Analyzing +4 | -1
If ee and e′e^{\prime} be the eccentricities of two conics S=0S=0 and S′=0S^{\prime}=0 and if e2+e′2=3e^2+e^{\prime 2}=3, then both SS and S′S^{\prime} can be
View answer and explanation
Correct option: A
Hyperbolas
Option A (Hyperbolas) is correct because for hyperbolas, the eccentricity is greater than 1, and it's possible to find two eccentricities that satisfy the given condition e2+e′2=3e^2 + e'^2 = 3.
84 Easy
84 of 298
Application +4 | -1
The equation x2−4y2−2x+16y−40=0x^2-4 y^2-2 x+16 y-40=0 represents
View answer and explanation
Correct option: C
A hyperbola
Option C (A hyperbola) is correct because the equation can be rearranged into the standard form of a hyperbola equation.
85 Easy
85 of 298
Application +4 | -1
The eccentricity of the hyperbola x2−y2=25x^2-y^2=25 is
View answer and explanation
Correct option: A
2\sqrt{2}
Option A (2\sqrt{2}) is correct because it is the calculated eccentricity of the given hyperbola after converting it to standard form.
86 Easy
86 of 298
Application +4 | -1
If a hyperbola passing through the origin has 3x−4y−1=03 x-4 y-1=0 and 4x−3y−6=04 x-3 y-6=0 as its asymptotes, then the equations of its transverse and conjugate axes, are
View answer and explanation
Correct option: C
x+y−5=0x+y-5=0 and x−y−1=0x-y-1=0
Option D (x+y−1=0x+y-1=0 and x−y−5=0x-y-5=0) is correct because these are the angle bisectors of the given asymptotes, representing the transverse and conjugate axes.
87 Easy
87 of 298
Application +4 | -1
The distance between the tangent to the hyperbola x24−y23=1,\frac{x^2}{4}-\frac{y^2}{3}=1, parallel to the line y=x+2y=x+2 is
View answer and explanation
Correct option: C
2\sqrt{2}
Option C (2\sqrt{2}) is correct because the distance between the tangents parallel to the given line is calculated using the formula for tangents to a hyperbola.
88 Easy
88 of 298
Application +4 | -1
The equation of the tangent to the curve 4x2−9y2=14 x^2-9 y^2=1 which is parallel to 4y=5x+74 y=5 x+7, is
View answer and explanation
Correct option: C
24y−30x=±16124 y-30 x=\pm \sqrt{161}
Option C (24y−30x=±16124 y-30 x=\pm \sqrt{161}) is correct because it represents the tangent lines to the hyperbola with the specified slope, derived using the tangent equation y=mx±a2m2−b2y = mx \pm \sqrt{a^2m^2 - b^2}.
89 Easy
89 of 298
Application +4 | -1
The coordinates of the point at which the line 3x+4y3 x+4 y =7=7 is a normal to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, are
View answer and explanation
Correct option: A
(7a23(a2+b2),7b24(a2+b2))\left(\frac{7 a^2}{3\left(a^2+b^2\right)}, \frac{7 b^2}{4\left(a^2+b^2\right)}\right)
Option A is correct because the coordinates are found by comparing the given normal equation with the standard normal equation of a hyperbola.
90 Easy
90 of 298
Application +4 | -1
The equation of the hyperbola, the length of whose latusrectum is 8 and eccentricity is 3/53 / \sqrt{5}, is
View answer and explanation
Correct option: B
4x2−5y2=1004 x^2-5 y^2=100
Option B (4x2−5y2=1004 x^2-5 y^2=100) is correct because it correctly uses the given latus rectum and eccentricity to derive the hyperbola equation.
Revision notes and key takeaways

Review HYPERBOLA in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 298 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - HYPERBOLA MCQs (All) 429 MCQs → ENGINEERING MATHEMATICS - HYPERBOLA MCQs (Hard) 13 MCQs → ENGINEERING MATHEMATICS - HYPERBOLA MCQs (Medium) 118 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

Your practice

0 / 298 answered · 298 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →