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JEE • MATHEMATICS • HYPERBOLA
Hyperbola practice · Easy
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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 241-270 of 298 questions
241
Easy
241 of 298
Application
+4 | -1
If circle intersect hyperbola at four points then
View answer and explanation
Correct option: B
Option B () is correct because it represents the sum of the roots of the quartic equation formed by the intersection of the circle and hyperbola.
242
Easy
242 of 298
Application
+4 | -1
View answer and explanation
Correct option: A
Option A () is correct because PF x PG equals which is the square of the semi-major axis CA.
243
Easy
243 of 298
Analyzing
+4 | -1
Let , where . Then, represents
View answer and explanation
Correct option: D
An ellipse whose eccentrieity is , where .
The correct answer is D. When r > 1, the equation represents an ellipse, and its eccentricity is calculated as .
244
Easy
244 of 298
Application
+4 | -1
The angle between the two asymptotes of the hyperbola is
View answer and explanation
Correct option: C
Option C () is correct because it represents the angle between the two asymptotes of the given hyperbola.
245
Easy
245 of 298
Application
+4 | -1
is a point on the hyperbola having vertices at . The equation of the normal to the hyperbola at is
View answer and explanation
Correct option: A
Option A () is correct because it represents the equation of the normal to the hyperbola at the given point P(10,16).
246
Easy
246 of 298
Analyzing
+4 | -1
Let and be two fixed points and , another point in the plane, moves in such a way that , where and are real constants. The locus of is
Which one in the options is not true?
Which one in the options is not true?
View answer and explanation
Correct option: D
A hyperbola if and
Option D (A hyperbola if and ) is not always true. The locus is not necessarily a hyperbola under these conditions.
247
Easy
247 of 298
Application
+4 | -1
The equations of the asymptotes of the hyperbola
are
are
View answer and explanation
Correct option: C
Option D (None of the above) is incorrect. Option A () is correct because the pair of asymptotes and the second-degree curve differ by a constant, and this constant can be found by setting the determinant to zero.
248
Easy
248 of 298
Application
+4 | -1
The equation of the hyperbola which is referred to its axes as axes of coordinate, and whose distance between the foci is 16 and eccentricity is , is
View answer and explanation
Correct option: B
Option B () is correct because given the distance between foci and eccentricity, we can determine 'a' and 'b' and construct the hyperbola's equation.
249
Easy
249 of 298
Understanding
+4 | -1
The angle between the asymptotes of is equal to
View answer and explanation
Correct option: A
Option A () is correct because the angle between the asymptotes of a hyperbola is given by this formula.
250
Easy
250 of 298
Analyzing
+4 | -1
The locus of the point of intersection of the straight lines and ( is a variable) is
View answer and explanation
Correct option: D
A hyperbola
Option D (A hyperbola) is correct because eliminating the variable λ results in the equation of a hyperbola.
251
Easy
251 of 298
Analyzing
+4 | -1
A series of hyperbola is drawn having a common transverse axis of length . Then the locus of a point on each hyperbola, such that its distance from the transverse axis is equal to its distance from an asymptote, is
View answer and explanation
Correct option: A
Option A () is correct because it represents the locus of point derived from the given condition that its distance from the transverse axis equals its distance from an asymptote.
252
Easy
252 of 298
Application
+4 | -1
The equation of the hyperbola referred to its axes as coordinate axes, whose latus rectum is 4 and eccentricity is 3 is
View answer and explanation
Correct option: C
Option C () is correct because given the latus rectum and eccentricity, the equation of the hyperbola can be derived as .
253
Easy
253 of 298
Application
+4 | -1
If the line be a tangent to the hyperbola then is equal to
View answer and explanation
Correct option: C
Option C () is correct because these are the values of for which the line is tangent to the hyperbola, found using the tangency condition.
254
Easy
254 of 298
Analyzing
+4 | -1
If the tangents drawn from a point on the hyperbola to the ellipse make angles and with the transverse axis of the hyperbola, then
View answer and explanation
Correct option: C
Option C () is correct because the product of the slopes of the tangents from a point on the hyperbola to the ellipse is 1.
255
Easy
255 of 298
Application
+4 | -1
Let be an ellipse, whose eccentricity is and the length of the latusrectum is . Then the square of the eccentricity of is:
View answer and explanation
Correct option: A
Option A () is correct because it is the square of the eccentricity of the hyperbola calculated using the given information about the ellipse.
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256
Easy
256 of 298
Application
+4 | -1
The eccentricity of the conic represented by , is
View answer and explanation
Correct option: B
Option B () is correct because the given equation represents a rectangular hyperbola, which always has an eccentricity of .
257
Easy
257 of 298
Application
+4 | -1
The locus of the poles of the chords of the hyperbola which subtend a right angle at its centre is
View answer and explanation
Correct option: C
Option C () is correct because the locus of the poles can be found by homogenizing the hyperbola equation using the polar equation and then applying the condition for perpendicular lines at the center.
258
Easy
258 of 298
Analyzing
+4 | -1
Eccentricity of the hyperbola is equal to
View answer and explanation
Correct option: A
The correct answer is because the angle between the asymptotes is 90 degrees, indicating a rectangular hyperbola, which has an eccentricity of .
259
Easy
259 of 298
Application
+4 | -1
The equation of the normal to the hyperbola at is
View answer and explanation
Correct option: D
Option D () is correct because the normal to a hyperbola at (-4,0) is a vertical line, and since the point lies on the x-axis, the normal is the x-axis itself.
260
Easy
260 of 298
Analyzing
+4 | -1
For the hyperbola , which of the following remains constant when varies?
View answer and explanation
Correct option: D
Abscissae of foci
Option D (Abscissae of foci) is correct because the x-coordinates of the foci remain constant at as varies.
261
Easy
261 of 298
Application
+4 | -1
What is the slope of the tangent line drawn to the hyperbola at the point ?
View answer and explanation
Correct option: B
Option B (-1/a) is correct because it is the derivative of the hyperbola equation evaluated at the specified point.
262
Easy
262 of 298
Application
+4 | -1
The locus of a point which is moving under the condition that the line is a tangent to the hyperbola is
View answer and explanation
Correct option: B
A hyperbola
Option B (A hyperbola) is correct because the condition for tangency to a hyperbola results in an equation that represents another hyperbola.
263
Easy
263 of 298
Application
+4 | -1
The locus of the point of intersection of the tangents to the hyperbola which are at right angles is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the locus is a circle which is not possible, so there is no pair of tangents that are at right angles.
264
Easy
264 of 298
Application
+4 | -1
The centre of the hyperbola , is
View answer and explanation
Correct option: A
Option A () is correct because the center is directly identified from the standard form of the hyperbola equation after completing the square.
265
Easy
265 of 298
Application
+4 | -1
Acceleration of a particle moving along a straight line is a function of velocity as . At , its velocity . Its velocity at s will be
View answer and explanation
Correct option: B
Option B () is correct because solving the differential equation with the given initial condition yields , and at t=3, v=25.
266
Easy
266 of 298
Application
+4 | -1
The co-ordinate of focus of
View answer and explanation
Correct option: A
Option A ((-6,1)) is correct because completing the square and transforming the equation into standard form reveals the center and allows calculation of the foci coordinates.
267
Easy
267 of 298
Application
+4 | -1
If the line touches the hyperbola , then
View answer and explanation
Correct option: B
Option B () is correct because it represents the condition for the line to be tangent to the hyperbola.
268
Easy
268 of 298
Analyzing
+4 | -1
The locus of the point of intersection of the tangents at the end-points of normal chords of the hyperbola , is
View answer and explanation
Correct option: B
Option B () is correct because it represents the locus of the point of intersection of tangents at the end-points of normal chords of the hyperbola.
269
Easy
269 of 298
Analyzing
+4 | -1
Let the foci of the ellipse and the hyperbola coincide. Then the length of the latus rectum of the hyperbola is:
View answer and explanation
Correct option: D
Option D () is correct because the length of the latus rectum is calculated using the formula after finding the value of by equating the foci of the ellipse and hyperbola.
270
Easy
270 of 298
Application
+4 | -1
The coordinates of the centre of the hyperbola
View answer and explanation
Correct option: A
Option A ((-1,0)) is correct because the center of the hyperbola is found by solving the partial derivatives of the equation with respect to x and y, setting them equal to zero.
Revision notes and key takeaways
Review HYPERBOLA in your textbook or class notes, then use this set to check your understanding.
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