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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Easy

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Showing 271-300 of 392 questions

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271 Easy
271 of 392
Application +4 | -1
If e1{e_1} and e2{e_2} are respectively the eccentricities of the ellipse x218+y24=1\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{4} = 1 and the hyperbola x29−y24=1,\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1, then the relation between e1{e_1} and e2{e_2} is
View answer and explanation
Correct option: C
2e12+e22=32e_1^2 + e_2^2 = 3
Option C (2e12+e22=32e_1^2 + e_2^2 = 3) is correct because it represents the correct relationship between the eccentricities of the given ellipse and hyperbola.
272 Easy
272 of 392
Application +4 | -1
If the polar with respect to y2=4ax{y^2} = 4ax touches the ellipse x2α2+y2β2=1,\frac{{{x^2}}}{{{\alpha ^2}}} + \frac{{{y^2}}}{{{\beta ^2}}} = 1, the locus of its pole is
View answer and explanation
Correct option: A
x2α2−y2(4a2α2/β2)=1\frac{{{x^2}}}{{{\alpha ^2}}} - \frac{{{y^2}}}{{\left( {4{a^2}{\alpha ^2}/{\beta ^2}} \right)}} = 1
Option A (x2α2−y2(4a2α2/β2)=1\frac{{{x^2}}}{{{\alpha ^2}}} - \frac{{{y^2}}}{{\left( {4{a^2}{\alpha ^2}/{\beta ^2}} \right)}} = 1) is correct because it represents the locus of the pole when the polar with respect to the parabola touches the given ellipse.
273 Easy
273 of 392
Application +4 | -1
The eccentricity of the conic 4x2+16y2−24x−32y=14{x^2} + 16{y^2} - 24x - 32y = 1 is
View answer and explanation
Correct option: C
32\frac{{\sqrt 3 }}{2}
The correct answer is C (32\frac{{\sqrt 3 }}{2}) because it represents the eccentricity of the given conic section after rewriting it in standard form and identifying the major and minor axes.
274 Easy
274 of 392
Application +4 | -1
The equation of normal at the point (0,3)\left( {0,3} \right) of the ellipse 9x2+5y2=45,9{x^2} + 5{y^2} = 45, is
View answer and explanation
Correct option: B
y−axisy - axis
Option B (y−axisy - axis) is correct because the normal at the point (0,3) of the ellipse is the y-axis.
275 Easy
275 of 392
Application +4 | -1
The locus of the middle points of chords of an ellipse which pass through a fixed point is
View answer and explanation
Correct option: A
An ellipse
Option A (An ellipse) is correct because the locus of the midpoints of chords passing through a fixed point in an ellipse is another ellipse.
276 Easy
276 of 392
Application +4 | -1
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0,53),(0,5\sqrt 3 ), then the length of its latus rectum is
View answer and explanation
Correct option: A
5
Option A (5) is correct because, using the given information about the ellipse and its foci, we can solve for the parameters and find the length of the latus rectum.
277 Easy
277 of 392
Analyzing +4 | -1
The tangent at a point P(θ)P(\theta) to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 cuts the auxiliary circle at points QQ and RR. If QRQ R subtends a right angle at the centre CC of the ellipse, then the eccentricity of the ellipse, is
View answer and explanation
Correct option: B
11+sin⁡2θ\frac{1}{\sqrt{1+\sin ^2 \theta}}
Option C (An ellipse) is correct because the locus of the point of intersection of tangents at P and Q to the ellipse is another ellipse.
278 Easy
278 of 392
Application +4 | -1
If equation of the ellipse is 2x2+3y2−8x+6y+5=02 x^2+3 y^2-8 x+6 y+5=0, then which of the following is true?
View answer and explanation
Correct option: C
Equation of auxiliary circle is x2+y2−4x+2y+2=0x^2+y^2-4 x+2 y+2=0
Option C (Equation of auxiliary circle is x2+y2−4x+2y+2=0x^2+y^2-4 x+2 y+2=0) is correct because after completing the squares and transforming the ellipse equation, the auxiliary circle's equation can be derived.
279 Easy
279 of 392
Application +4 | -1
If CC is the centre of the ellipse 9x2+16y2=1449{x^2} + 16{y^2} = 144 and SS is one focus. The ratio of CSCS to major axis, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the ratio of CS to the major axis is calculated as 7:8\sqrt{7}:8, which is not among the provided options.
280 Easy
280 of 392
Understanding +4 | -1
PP is any point on the ellipse 81x2+144y2=194481{x^2} + 144{y^2} = 1944, whose foci are SS and S′S'. Then, SP+S′PSP + S'P equals
View answer and explanation
Correct option: B
464\sqrt 6
Option B (464\sqrt 6) is correct because SP+S′PSP + S'P equals to 2a2a which is 464\sqrt{6} based on the given ellipse equation.
281 Easy
281 of 392
Analyzing +4 | -1
If the ellipse x24+y2=1\frac{x^2}{4}+y^2=1 meets the ellipse x2+y2a2=1x^2+\frac{y^2}{a^2}=1 in four distinct points and a=b2−5b+7a=b^2-5 b+7, then bb belongs to
View answer and explanation
Correct option: B
(−∞,2)∪(3,∞)(-\infty, 2) \cup(3, \infty)
Option C ((2,3)(2,3)) is correct because the condition for four distinct intersection points implies that a>1a > 1, leading to the inequality b2−5b+6>0b^2 - 5b + 6 > 0, which gives the solution b in (2,3).
282 Easy
282 of 392
Application +4 | -1
The point (4,−3)(4,-3) w.r.t. the ellipse 4x2+5y2=14 x^2+5 y^2=1
View answer and explanation
Correct option: C
Is outside the curve
Option C (Is outside the curve) is correct because substituting the point into the ellipse equation results in a value greater than 1, indicating it lies outside.
283 Easy
283 of 392
Application +4 | -1
For the ellipse 3x2+4y2=123 x^2+4 y^2=12, the length of latus rectum is
View answer and explanation
Correct option: B
3
Option B (3) is correct because after converting the ellipse equation to standard form, the latus rectum is calculated as 2b2a\frac{{2b^2}}{{a}}.
284 Easy
284 of 392
Understanding +4 | -1
The equation 2x2+3y2=302 x^2+3 y^2=30 represents
View answer and explanation
Correct option: B
An ellipse
Option B (An ellipse) is correct because the equation can be rewritten in the standard form of an ellipse equation.
285 Easy
285 of 392
Application +4 | -1
If SS and S′S^{\prime} are two foci of the ellipse 16x2+25y2=40016 x^2+25 y^2=400 and PSQ is a focal chord such that SP=16S P=16, then SQ=S Q=
View answer and explanation
Correct option: D
74/974 / 9
Option D (74/974 / 9) is correct because it is derived using the harmonic mean of focal segments and the property that the sum of distances from a point on the ellipse to the foci equals the major axis.

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286 Easy
286 of 392
Application +4 | -1
Area of quadrilateral formed by joining the focii of the ellipse, x216+y212=1\frac{x^2}{16}+\frac{y^2}{12}=1 and x212+y216=1\frac{x^2}{12}+\frac{y^2}{16}=1 is (in sq. units)
View answer and explanation
Correct option: D
8
Option D (2xyℓn(xy)=x2+C2 x y \ell n(x y)=x^2+C) is correct because by rearranging the differential equation and integrating, we arrive at the given solution.
287 Easy
287 of 392
Application +4 | -1
If y=xy=x and 3y+2x=03 y+2 x=0 are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is
View answer and explanation
Correct option: B
1/31 / \sqrt{3}
Option B (1/31 / \sqrt{3}) is correct because it's derived from the relationship between the slopes of conjugate diameters and the ellipse's eccentricity.
288 Easy
288 of 392
Application +4 | -1
In triangle ABC\mathrm{ABC} given the base BC\mathrm{BC} of the triangle and if ∠C−∠B=K\angle C-\angle{B}=K, a constant. Then the locus of the vertex A\mathrm{A} is
View answer and explanation
Correct option: A
hyperbola
Option A (hyperbola) is correct because the locus of the vertex A, given the base BC and a constant difference between angles C and B, results in the equation of a hyperbola.
289 Easy
289 of 392
Application +4 | -1
Equation of the ellipse whose axes are the axes of co-ordinates and which passes through the point (−3,1)(-3,1) and has eccentricity 25\sqrt{\frac{2}{5}}, is
View answer and explanation
Correct option: D
5x2+3y2−32=05 x^2+3 y^2-32=0
Option D (5x2+3y2−32=05 x^2+3 y^2-32=0) is correct because it is the equation of the ellipse that passes through the given point and has the given eccentricity.
290 Easy
290 of 392
Application +4 | -1
The locus of the foot of the perpendicular drawn from the centre of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 on any tangent, is
View answer and explanation
Correct option: C
(x2+y2)2=a2x2+b2y2\left(x^2+y^2\right)^2=a^2 x^2+b^2 y^2
Option C ((x2+y2)2=a2x2+b2y2(x^2+y^2)^2=a^2 x^2+b^2 y^2) is correct because it represents the locus of the foot of the perpendicular from the center of the ellipse on any tangent.
291 Easy
291 of 392
Application +4 | -1
If the foci of the ellipse x216+y2b2=1\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1 coincide with the foci of the hyperbola x2144−y281=125\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}, then b2b^{2} is equal to
View answer and explanation
Correct option: C
7
Option C (7) is correct because the value of b2b^2 is calculated by equating the foci of the ellipse and hyperbola.
292 Easy
292 of 392
Application +4 | -1
The line 12xcos⁡θ + 5ysin⁡θ = 6012x\cos \theta \, + \,5y\sin \theta \, = \,60 is tangent to which of the following curves?
View answer and explanation
Correct option: B
144x2+25y2=3600144{x^2} + 25{y^2} = 3600
Option B (144x2+25y2=3600144{x^2} + 25{y^2} = 3600) is correct because the given line is tangent to the ellipse x225+y2144=1\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{144}} = 1, which is equivalent to the given equation.
293 Easy
293 of 392
Understanding +4 | -1
If AA and BB are two fixed points and PP is a variable point such that PA+PB=4,PA + PB = 4, the locus of PP is
View answer and explanation
Correct option: B
An ellipse
Option B (An ellipse) is correct because the sum of distances from two fixed points being constant defines an ellipse.
294 Easy
294 of 392
Application +4 | -1
The eccentricity of the ellipse 9x2+5y2−30y=09{x^2} + 5{y^2} - 30y = 0 is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because the eccentricity of the ellipse is calculated using the formula e=1−a2b2e = \sqrt{1 - \frac{a^2}{b^2}} after rewriting the given equation in standard form.
295 Easy
295 of 392
Application +4 | -1
The line y=x+1y = x + 1 meets the ellipse x24+y22=1\frac{{{x^2}}}{4} + \frac{{{y^2}}}{2} = 1 at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2{(3r)^2} is equal to:
View answer and explanation
Correct option: A
20
Option B (9) is correct because it is the value of 12m2, where m is the slope of the common tangent to the given ellipse and circle, calculated by equating the tangent equations.
296 Easy
296 of 392
Application +4 | -1
If α\alpha and β\beta are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse, is
View answer and explanation
Correct option: D
sin⁡α+sin⁡βsin⁡(α+β)\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}
Option D (sin⁡α+sin⁡βsin⁡(α+β)\frac{{\sin \alpha + \sin \beta}}{{\sin (\alpha + \beta)}}) is correct because the eccentricity is derived using the equation of a focal chord and the properties of eccentric angles.
297 Easy
297 of 392
Understanding +4 | -1
The sum of the focal distances from any point on the ellipse 9x2+16y2=1449{x^2} + 16{y^2} = 144 is
View answer and explanation
Correct option: C
8
Option C (8) is correct because the sum of the focal distances from any point on the ellipse is equal to the length of its major axis (2a).
298 Easy
298 of 392
Application +4 | -1
Eccentricity of the ellipse whose latus rectum is half of its major axis is:
View answer and explanation
Correct option: B
12\frac{1}{\sqrt{2}}
Option B (12\frac{1}{\sqrt{2}}) is correct because the eccentricity is calculated using the relationship between the latus rectum and the major axis.
299 Easy
299 of 392
Application +4 | -1
The length of the latusrectum of the ellipse 5x2+9y2=455{x^2} + 9{y^2} = 45 is
View answer and explanation
Correct option: B
103\frac{{10}}{3}
Option B (103\frac{{10}}{3}) is correct because the length of the latus rectum of an ellipse is given by 2b2a\frac{{2{b^2}}}{a}, where a and b are the semi-major and semi-minor axes respectively.
300 Easy
300 of 392
Application +4 | -1
On the ellipse x24+y29=1\frac{x^2}{4}+\frac{y^2}{9}=1, one of the points of which the normals are parallel to the line 2x−y=12 x-y=1, is
View answer and explanation
Correct option: C
(210,910)\left(\frac{2}{\sqrt{10}}, \frac{9}{\sqrt{10}}\right)
Option C ((210,910)\left(\frac{2}{\sqrt{10}}, \frac{9}{\sqrt{10}}\right)) is correct because it represents a point on the ellipse where the normal's slope matches that of the given line.
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