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JEE • MATHEMATICS • HYPERBOLA
Hyperbola practice · Mixed difficulty
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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.
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Showing 301-330 of 429 questions
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301
Easy
301 of 429
Application
+4 | -1
If represents a hyperbola, then
View answer and explanation
Correct option: D
or
Option D ( or ) is correct because for the equation to represent a hyperbola, either and or and must be true.
302
Easy
302 of 429
Application
+4 | -1
The eccentricity of the hyperbola is
View answer and explanation
Correct option: B
Option B () is correct because the given hyperbola is a rectangular hyperbola (since a=b), and the eccentricity of any rectangular hyperbola is always .
303
Easy
303 of 429
Application
+4 | -1
If the normal at to the rectangular hyperbola meets the axes of and in and respectively and is the centre of the hyperbola, then
View answer and explanation
Correct option: C
Option C () is correct because the distance between the points where the normal intersects the x and y axes is twice the distance from the center of the hyperbola to point P.
304
Easy
304 of 429
Application
+4 | -1
If the origin is shifted to the point and the axes are rotated through an angle in a clockwise direction, so that the equation of the given hyperbola changes to the standard form , then is
View answer and explanation
Correct option: B
Option B () is correct because the angle of rotation is the angle whose tangent is the slope of the transverse axis.
305
Easy
305 of 429
Application
+4 | -1
If , then the chord joining the points and for the hyperbola passes through
View answer and explanation
Correct option: B
centre
Option B (centre) is correct because when , the equation of the chord simplifies to one that always passes through the origin (0,0), which is the center of the hyperbola.
306
Medium
306 of 429
Application
+4 | -1
If the chord of contact of tangents from a point to the hyperbola subtends a right angle at the centre, then the locus of is
View answer and explanation
Correct option: B
An ellipse
Option B is correct because the locus of P is an ellipse, derived from the condition that the chord of contact subtends a right angle at the center.
307
Easy
307 of 429
Application
+4 | -1
The hyperbola does not have mutually tangents, then possible values of eccentricities are
a) b) c) d)
a) b) c) d)
View answer and explanation
Correct option: C
Only a,c are correct
Option C (Only a,c are correct) is correct because the condition for the hyperbola not to have mutually perpendicular tangents is that e > , and only and satisfy this.
308
Medium
308 of 429
Application
+4 | -1
Find the equation of the hyperbola whose directrix is , focus is and eccentricity is .
View answer and explanation
Correct option: A
Option A () is correct because it is the equation of the hyperbola derived using the focus-directrix definition and the given eccentricity.
309
Easy
309 of 429
Application
+4 | -1
The eccentricity of the hyperbola is
View answer and explanation
Correct option: C
Option C () is correct because it is the calculated eccentricity of the given hyperbola after converting it to standard form.
310
Medium
310 of 429
Application
+4 | -1
If the eccentricity of the hyperbola is times the eccentricity of the ellipse , then a value of is
View answer and explanation
Correct option: B
Option B () is correct because it is the value of that satisfies the given condition relating the eccentricities of the hyperbola and the ellipse.
311
Easy
311 of 429
Analyzing
+4 | -1
Consider a branch of the hyperbola with vertex at the point . Let be one of the end points of its latusrectum. If is the focus of the hyperbola nearest to the point , then the area of the triangle , is
View answer and explanation
Correct option: B
Option B () is correct because it is the area of the triangle calculated using the coordinates of the vertex, focus, and endpoint of the latus rectum.
312
Medium
312 of 429
Application
+4 | -1
The locus of a point moving under the condition that the line is a tangent to the hyperbola , is
View answer and explanation
Correct option: A
A hyperbola
Option A (A hyperbola) is correct because the locus of the point is derived from the condition of tangency to a hyperbola, resulting in another hyperbola.
313
Medium
313 of 429
Analyzing
+4 | -1
The condition that a straight line with slope will be normal to parabola as well as a tangent to rectangular hyperbola is
View answer and explanation
Correct option: D
Option D () is correct because it represents the condition derived by equating the normal to the parabola and the tangent to the rectangular hyperbola.
314
Easy
314 of 429
Application
+4 | -1
The equation of the normal at the point of the curve is
View answer and explanation
Correct option: C
The correct answer is because this is the standard form of the normal equation to a hyperbola at the given point.
315
Medium
315 of 429
Application
+4 | -1
Tangent to hyperbola at point intersects the -axis at and the -axis at . Normal to hyperbola at intersects the -axis at and the -axis at . If the area of the triangles and are and respectively then , is equal to
View answer and explanation
Correct option: B
Option B () is correct because, after finding the coordinates of T, T', N, and N', and calculating the areas of the triangles PNT and PN'T', the expression simplifies to .
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316
Easy
316 of 429
Understanding
+4 | -1
The eccentricity of the curve is
View answer and explanation
Correct option: D
Option D () is correct because the eccentricity of a rectangular hyperbola (where a=b) is always .
317
Easy
317 of 429
Application
+4 | -1
The equation represents
View answer and explanation
Correct option: C
A hyperbola
Option C (A hyperbola) is correct because the discriminant () is positive, indicating a hyperbola.
318
Easy
318 of 429
Application
+4 | -1
The eccentricity of the conic defined by is , then
View answer and explanation
Correct option: D
5
Option D (5) is correct because the eccentricity of the hyperbola is calculated using the given foci and transverse axis length, resulting in K = 5/3, and 3K = 5.
319
Easy
319 of 429
Application
+4 | -1
If the slope of a tangent to the hyperbola is , then the eccentricity of the hyperbola lies in the interval.
View answer and explanation
Correct option: C
Option C () is correct because the condition for the tangent to exist implies lies in the interval .
320
Easy
320 of 429
Analyzing
+4 | -1
From a point on the line (parameter), tangents are drawn to the hyperbola such that chords of contact pass through a fixed point . Then, is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because the chord of contact passes through a fixed point, and the ratio can be determined from the equation of the chord of contact.
321
Easy
321 of 429
Application
+4 | -1
The eccentricity of the hyperbola is
View answer and explanation
Correct option: C
Option C () is correct because the eccentricity of the hyperbola is calculated using the formula .
322
Medium
322 of 429
Analyzing
+4 | -1
If the circle intersects the hyperbola in four points and , then
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the sum of the x-coordinates and y-coordinates of the intersection points are zero, and the product of the x-coordinates equals the product of the y-coordinates, both equal to .
323
Medium
323 of 429
Application
+4 | -1
A common tangent to and is
View answer and explanation
Correct option: B
Option B () is correct because it satisfies the tangency conditions for both the hyperbola and the circle.
324
Easy
324 of 429
Application
+4 | -1
The latusractum of the hyperbola is
View answer and explanation
Correct option: A
Option A (9/2) is correct because it is the result of calculating the latus rectum using the formula 2b2/a after rewriting the hyperbola equation in standard form.
325
Hard
325 of 429
Application
+4 | -1
If the angle between the asymptotes of the hyperbola is and is its eccentricity then is equal to
View answer and explanation
Correct option: D
Option D () is correct because it correctly relates the angle between asymptotes and the eccentricity of the hyperbola.
326
Hard
326 of 429
Analyzing
+4 | -1
Let and where be two points on the hyperbola . If is the point of intersection of normals at and , then is equal to
View answer and explanation
Correct option: D
Option D () is correct because it is the y-coordinate of the intersection point of the normals at P and Q, derived using the given condition θ + φ = π/2.
327
Easy
327 of 429
Application
+4 | -1
is any point on the hyperbola . If and are the foci of the hyperbola and , where is the origin, then is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by substituting the given hyperbola equation and simplifying, we find that equals 1.
328
Easy
328 of 429
Application
+4 | -1
If a hyperbola passes through the point and it has vertices at then the equation of the normal to it at is
View answer and explanation
Correct option: C
Option B () is correct because it is the length of the perpendicular from the origin to the normal of the curve at the given point.
329
Easy
329 of 429
Application
+4 | -1
The equation of the tangent to the hyperbola at , is
View answer and explanation
Correct option: B
Option B () is correct because it is the equation of the tangent to the hyperbola at the given point, found by substituting the point into the hyperbola's equation.
330
Easy
330 of 429
Application
+4 | -1
If the normal at the point '' to the rectangular hyperbola meets it again at the point '', then
View answer and explanation
Correct option: A
Option A () is correct because it represents the relationship between the parameters of the points where the normal to a rectangular hyperbola intersects the curve again.
Revision notes and key takeaways
Review HYPERBOLA in your textbook or class notes, then use this set to check your understanding.
Showing 301-330 of 429 questions
