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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Medium

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Showing 121-150 of 187 questions

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121 Medium
121 of 187
Application +4 | -1
Suppose xx and yy are real numbers and that x2+9y2−4x+6y+4=0x^2+9 y^2-4 x+6 y+4=0, then the maximum value of (4x−9y)(4 x-9 y).
View answer and explanation
Correct option: C
16
Option C (16) is correct because after completing the square and converting the equation to the standard form of an ellipse, the maximum value of 4x - 9y is found to be 16.
122 Medium
122 of 187
Application +4 | -1
If the length of the latusrectum of the ellipse x2tan⁡2θ+y2sec⁡2θ=1{x^2}{\tan ^2}\theta + {y^2}{\sec ^2}\theta = 1 is 12\frac{1}{2}, then θ=\theta =
View answer and explanation
Correct option: A
π/12,5π/12\pi /12,5\pi /12
Option A (π/12,5π/12\pi /12,5\pi /12) is correct because these are the values of θ\theta that satisfy the given condition for the length of the latus rectum.
123 Medium
123 of 187
Analyzing +4 | -1
If QQ is the point on the auxiliary circle corresponding to a point PP on an ellipse.
Then, the normals at PP and QQ meet on
View answer and explanation
Correct option: A
A fixed circle
Option A (A fixed circle) is correct because the normals at P and Q meet on a fixed circle.
124 Medium
124 of 187
Analyzing +4 | -1
A tangent at any point to the ellipse 4x2+9y2=364{x^2} + 9{y^2} = 36 is cut by the tangent at the extremities of the major axis at TT and T′T'. The circle on TT′TT' as diameter passes through the point
View answer and explanation
Correct option: B
(5,0)\left( {\sqrt 5 ,0} \right)
Option B ((5,0)\left( {\sqrt 5 ,0} \right)) is correct because the circle on TT' as diameter passes through this point, as derived from the equation of the circle.
125 Medium
125 of 187
Application +4 | -1
The eccentricity of the ellipse 4x2+9y2=364 x^2+9 y^2=36, is
View answer and explanation
Correct option: C
53\frac{\sqrt{5}}{3}
Option C (53\frac{{\sqrt{{5}}}}{{3}}) is correct because the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}} after converting the equation to standard form.
126 Medium
126 of 187
Analyzing +4 | -1
Let the tangent and normal at the point (33,1)(3 \sqrt{3}, 1) on the ellipse x236+y24=1\frac{x^{2}}{36}+\frac{y^{2}}{4}=1 meet the yy-axis at the points AA and BB respectively. Let the circle CC be drawn taking ABA B as a diameter and the line x=25x=2 \sqrt{5} intersect CC at the points PP and QQ. If the tangents at the points PP and QQ on the circle intersect at the point (α,β)(\alpha, \beta), then α2−β2\alpha^{2}-\beta^{2} is equal to
View answer and explanation
Correct option: C
3045\frac{304}{5}
The correct answer is C (3045\frac{304}{5}) because it is the value of α2−β2\alpha^2 - \beta^2 after finding the intersection point of tangents to the circle constructed with the segment AB as diameter.
127 Medium
127 of 187
Analyzing +4 | -1
If the normals at the points A,B,CA, B, C and DD of an ellipse meet in a point OO, then SA⋅SB⋅SC⋅SDS A \cdot S B \cdot S C \cdot S D is equal to (where, SS is one of the foci)
View answer and explanation
Correct option: A
b2e2⋅SO2\frac{b^2}{e^2} \cdot S O^2
Option A (b2e2⋅SO2\frac{b^2}{e^2} \cdot S O^2) is correct because after substituting and simplifying the expression for the product of distances from the foci to the points, it equals b2e2⋅SO2\frac{b^2}{e^2} \cdot S O^2.
128 Medium
128 of 187
Analyzing +4 | -1
If lx+my+n=0lx + my + n = 0 represents a chord of the ellipse b2x2+a2y2=a2b2{b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2} whose eccentric angles differ by 900{90^0}, then
View answer and explanation
Correct option: C
a2l2+b2m2=2n2{a^2}{l^2} + {b^2}{m^2} = 2{n^2}
Option A (Focus) is correct because the normal chord at (x1, x1) subtends a right angle at the focus of the parabola.
129 Medium
129 of 187
Application +4 | -1
The eccentricity of the conic 4x2+16y2−24x−32y=14 x^2+16 y^2-24 x-32 y=1 is
View answer and explanation
Correct option: A
32\frac{\sqrt{3}}{2}
Option A (32\frac{{\sqrt{3}}}{{2}}) is correct because, after rewriting the equation in standard form, the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}}.
130 Medium
130 of 187
Analyzing +4 | -1
An ellipse whose focii are (2,4)(2,4) and (14,9)(14,9) and touches x-axis, then its eccentricity is
View answer and explanation
Correct option: C
13313\frac{13}{\sqrt{313}}
Option C (13313\frac{13}{\sqrt{313}}) is correct because it is derived from the geometric properties of the ellipse and the given foci and tangency condition.
131 Medium
131 of 187
Application +4 | -1
Coordinates of the foci of the ellipse 2x2+3y2−4x−12y+13=02 x^2+3 y^2-4 x-12 y+13=0 are
View answer and explanation
Correct option: A
(1±16,2)\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)
Option A ((1±16,2)\left(1 \pm \frac{{1}}{{\sqrt{{6}}}}, 2\right)) is correct because it is derived by completing the square, finding the ellipse parameters, and calculating the foci coordinates.
132 Medium
132 of 187
Analyzing +4 | -1
If the normal at any point PP of the ellipse x2a2+y2b2=1(a>b>0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0) meets the major and minor axes at G{G} and G′G^{\prime} respectively and OF{OF} is the perpendicular drawn from centre O{O} to this normal, then PF⋅PGPF \cdot PG is equal to
View answer and explanation
Correct option: A
b2{b}^2
The correct answer is A (b2{b}^2). By finding the coordinates of G and using the perpendicular distance formula, we can derive that PF⋅PG=b2PF \cdot PG = b^2.
133 Medium
133 of 187
Application +4 | -1
If the tangents are drawn to the ellipse x2+2y2=2{x^2} + 2{y^2} = 2, then the locus of the mid point of the intercept made by the tangents between the coordinate axes is
View answer and explanation
Correct option: A
12x2+14y2=1\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1
Option A (12x2+14y2=1\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1) is correct because it represents the locus of the midpoint of the intercept made by the tangents between the coordinate axes, derived using the equation of the tangent to the ellipse.
134 Medium
134 of 187
Analyzing +4 | -1
If the line xcos⁡α+ysin⁡α=px\cos \mathrm{\alpha } + y\sin \mathrm{\alpha } = p be normal to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, then
View answer and explanation
Correct option: D
p2(a2sec⁡2α+b2cosec2α)=(a2−b2)2{p^2}\left( {{a^2}{{\sec }^2}\mathrm{\alpha } + {b^2}\mathrm{cose}{\mathrm{c}^2}\mathrm{\alpha }} \right) = {\left( {{a^2} - {b^2}} \right)^2}
Option D (p2(a2sec⁡2α+b2cosec2α)=(a2−b2)2{p^2}\left( {{a^2}{{\sec }^2}\alpha + {b^2}\mathrm{cose}{\mathrm{c}^2}\alpha } \right) = {\left( {{a^2} - {b^2}} \right)^2}) is correct because it is derived by equating the normal equation of the ellipse with the given line equation.
135 Medium
135 of 187
Analyzing +4 | -1
Let the ellipse E:x2+9y2=9E: x^2+9 y^2=9 intersect the positive xx - and yy-axes at the points AA and BB respectively. Let the major axis of EE be a diameter of the circle CC. Let the line passing through AA and BB meet the circle CC at the point PP. If the area of the triangle with vertices A,PA, P and the origin OO is mn\frac{m}{n}, where mm and nn are coprime, then m−nm-n is equal to
View answer and explanation
Correct option: C
17
The correct answer is C (17) because the area of the triangle AOP is calculated using the intersection of the ellipse, circle, and line equations, resulting in an area of 27/10, and thus m-n = 27-10 = 17.
136 Medium
136 of 187
Application +4 | -1
If mm is the slope of a common tangent to the curves x216+y29=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1 and  x2+y2=12,\,{{\text{x}}^2} + {y^2} = 12, then 12m2{\text{12}}{m^2} is equal to:
View answer and explanation
Correct option: B
9
Option B (9) is correct because after finding the condition for a common tangent and solving for m2m^2, we find that 12m2=912m^2 = 9.
137 Medium
137 of 187
Application +4 | -1
The distance of the point 'θ\theta' on the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 from a focus is
View answer and explanation
Correct option: C
a(1+ecos⁡θ)a\left( {1 + e\cos \theta } \right)
Option C, a(1+ecos⁡θ)a(1 + e\cos \theta), is correct because the focal distance is derived using the ellipse equation and the definition of eccentricity.
138 Medium
138 of 187
Application +4 | -1
The locus of the foot of the perpendicular from the foci on any tangent to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, is
View answer and explanation
Correct option: B
x2+y2=a2x^2+y^2=a^2
Option B (x2+y2=a2x^2+y^2=a^2) is correct because it represents the locus of the foot of the perpendicular from the foci on any tangent to the ellipse, which is the auxiliary circle.
139 Medium
139 of 187
Application +4 | -1
The equation of the normal to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at the positive end of the latusrectum, is
View answer and explanation
Correct option: B
x−ey−e3a=0x-e y-e^3 a=0
Option B (x−ey−e3a=0x-e y-e^3 a=0) is correct because it represents the equation of the normal to the ellipse at the positive end of the latus rectum after substituting the coordinates and simplifying.
140 Medium
140 of 187
Application +4 | -1
The equation of the tangents which are drawn at the ends of the major axis of the ellipse 9x2+5y2−30y=09 x^2+5 y^2-30 y=0 are
View answer and explanation
Correct option: C
y=0,y=6y=0, y=6
Option C (y=0,y=6y=0, y=6) is correct because these lines represent the tangents at the ends of the major axis of the given ellipse.
141 Medium
141 of 187
Analyzing +4 | -1
If y=mx−(a2−b2)ma2+b2m2y = mx - \frac{{\left( {{a^2} - {b^2}} \right)m}}{{\sqrt {{a^2} + {b^2}{m^2}} }} is normal to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 for all values of mm belonging to
View answer and explanation
Correct option: C
R
Option C (R) is correct because the condition for the line to be normal to the ellipse holds for all real values of m, which is the slope of the normal.
142 Medium
142 of 187
Analyzing +4 | -1
Ellipse 2x2+y2=12 x^2+y^2=1 and the line y=mxy=m x meet at a point PP in the first quadrant. Normal to the ellipse at PP meets xx-axis at (−132,0)\left(-\frac{1}{3 \sqrt{2}}, 0\right) and yy-axis at (0,β)(0, \beta), then ∣β∣=|\beta|=
View answer and explanation
Correct option: C
23\frac{\sqrt{2}}{3}
Option C (23\frac{\sqrt{2}}{3}) is correct because it is derived from the equation of the normal to the ellipse and the given x-intercept, combined with the ellipse equation to solve for the y-coordinate of point P.
143 Medium
143 of 187
Analyzing +4 | -1
If the curves x24+y2=1\frac{x^2}{4}+y^2=1 and x2a2+y2=1\frac{x^2}{a^2}+y^2=1 for a suitable value of a cut on four concyclic points, the equation of the circle passing through these four points is
View answer and explanation
Correct option: D
x2+y2=1x^2+y^2=1
Option D (x2+y2=1x^2+y^2=1) is correct because the equation of the circle passing through the intersection points of the two ellipses is found by setting up a general equation and solving for the parameter that makes it a circle.
144 Medium
144 of 187
Analyzing +4 | -1
PCP′P C P^{\prime} and DCD′D C D^{\prime} are conjugate diameters of an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and θ\theta is the eccentric angle of PP. Then, eccentric angle of the point where the circle PP′DP P^{\prime} D again cuts the ellipse
View answer and explanation
Correct option: D
π2−3θ\frac{\pi}{2}-3 \theta
Option D (π2−3θ\frac{\pi}{2}-3 \theta) is correct because it is derived from the property that the sum of eccentric angles where a circle cuts an ellipse is an even multiple of π\pi.
145 Medium
145 of 187
Application +4 | -1
Equation of a circle described on the Latusrectum of the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1 as a diameter with centre on the + ve xx-axis is
View answer and explanation
Correct option: C
25x2+25y2−150x−31=025 x^2+25 y^2-150 x-31=0
Option C (25x2+25y2−150x−31=025 x^2+25 y^2-150 x-31=0) is correct because the center of the circle is at the focus of the ellipse (3,0) and the radius is the semi-latus rectum (16/5).
146 Medium
146 of 187
Understanding +4 | -1
Sum of the focal distance of any point on the ellipse x2/a2+y2/b2=1x^2 / a^2+y^2 / b^2=1 is equal to the length of the
View answer and explanation
Correct option: A
Major axis
Option A (Major axis) is correct because the sum of the focal distances of any point on an ellipse is constant and equal to the length of the major axis (2a).
147 Medium
147 of 187
Understanding +4 | -1
If θ\mathrm{\theta } and ϕ\phi are eccentric angle of the ends of a pair of conjugate diameters of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, then θ−ϕ\mathrm{\theta } - \phi is equal to
View answer and explanation
Correct option: A
± π2\pm \,\frac{\pi }{2}
Option A (10) is correct because the sum of the focal distances of any point on the ellipse is equal to the length of the major axis, which is 2a=102a = 10.
148 Medium
148 of 187
Analyzing +4 | -1
If the normal at point PP on the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 meets the axes in RR and SS respectively, then PR:RSPR:RS is equal to
View answer and explanation
Correct option: C
b2:a2{b^2}:{a^2}
Option C (b2:a2{b^2}:{a^2}) is correct because it represents the ratio of PR to RS, derived from the equation of the normal to the ellipse and the coordinates of the points where the normal intersects the axes.
149 Medium
149 of 187
Analyzing +4 | -1
If the normal at any point P{P} on an ellipse of eccentricity 'ee' meets the major axis and minor axis at TT and GG respectively, then PT:PGPT : PG is equal to
View answer and explanation
Correct option: C
(1−e2):1\left(1-e^{2}\right): 1
Option C (\left(1-e2\right): 1) is correct because it represents the ratio of PT to PG, which is equal to b2/a2, and b2/a2 = 1-e2 for an ellipse.
150 Medium
150 of 187
Application +4 | -1
If the eccentricity of the two ellipses x2169+y225=1\frac{x^2}{169}+\frac{y^2}{25}=1 and x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 are equal, then the value of a/ba / b is
View answer and explanation
Correct option: C
13/513 / 5
Option C (13/513 / 5) is correct because the eccentricities are equated and the ratio a/ba/b is calculated.
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