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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Medium

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Showing 61-90 of 187 questions
61 Medium
61 of 187
Application +4 | -1
Let PP be a point on the ellipse x2a2+y2b2=1,\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, of eccentricity ee. If A,A′A,A' are the vertices and S,S′S,S' are the foci of the ellipse, then Area ΔPSS′:Area  ΔAPA′=\Delta PSS':\mathrm{Area}\,\,\Delta APA' =
View answer and explanation
Correct option: C
e:1e:1
Option C (e:1e:1) is correct because the ratio of the areas of the two triangles is equal to the eccentricity of the ellipse.
62 Medium
62 of 187
Analyzing +4 | -1
If θ1,θ2,θ3,θ4\theta_1, \theta_2, \theta_3, \theta_4 are eccentric angles of 4 concyclic points on Ellipse x24+y29=1\frac{x^2}{4}+\frac{y^2}{9}=1 Then cos⁡(θ1+θ2+θ3+θ44)\cos \left(\frac{\theta_1+\theta_2+\theta_3+\theta_4}{4}\right) cannot be
View answer and explanation
Correct option: D
1/21 / 2
Option D (1/21 / 2) is correct because the sum of the eccentric angles of four concyclic points on an ellipse is an even multiple of π\pi, and therefore the cosine of their average cannot be 1/21/2.
63 Medium
63 of 187
Analyzing +4 | -1
The locus of the poles of the chords of the ellipse subtending a right angle at its centre is
View answer and explanation
Correct option: C
x2a4+y2b4=1a2+1b2\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}
Option C (x2a4+y2b4=1a2+1b2\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}) is correct because it represents the locus of the poles of chords subtending a right angle at the center of the ellipse.
64 Medium
64 of 187
Analyzing +4 | -1
If the normals to x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at the ends of the chords l1x+m1y=1l_1 x+m_1 y=1 and l2x+m2y=1l_2 x+m_2 y=1 are concurrent, then
View answer and explanation
Correct option: B
a2l1l2=b2m1m2=−1a^2 l_1 l_2=b^2 m_1 m_2=-1
Option B is correct because it represents the condition for concurrency of normals at the ends of the given chords.
65 Medium
65 of 187
Application +4 | -1
Tangent are drawn from any point on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 to the circle x2+y2=r2x^2+y^2=r^2, then the chord of contact is a tangent to the ellipse, is
View answer and explanation
Correct option: B
a2x2+b2y2=r4a^2 x^2+b^2 y^2=r^4
Option B (a2x2+b2y2=r4a^2 x^2+b^2 y^2=r^4) is correct because it represents the condition for the chord of contact to be tangent to another ellipse.
66 Medium
66 of 187
Analyzing +4 | -1
Three points A,B,CA,B,C are taken on the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 with eccentric angles θ,θ+α and θ+2α\mathrm{\theta },\mathrm{\theta } + \mathrm{\alpha }\,\mathrm{and }\,\mathrm{\theta } + 2\mathrm{\alpha }, then
View answer and explanation
Correct option: D
The maximum value of area is 334ab\frac{{3\sqrt 3 }}{4}ab
Option D (The maximum value of area is 334ab\frac{{3\sqrt 3 }}{4}ab) is correct because the area of the triangle formed by the points on the ellipse is maximized when cos⁡α=−12\cos \alpha = -\frac{1}{2}, resulting in the maximum area of 334ab\frac{{3\sqrt 3 }}{4}ab.
67 Medium
67 of 187
Application +4 | -1
Equation of the latusrectum of the ellipse 9x2+4y2−18x−8y−23=0  9{x^2} + 4{y^2} - 18x - 8y - 23 = 0\; are
View answer and explanation
Correct option: C
y=1±5y = 1 \pm \sqrt 5
Option C (y=1±5y = 1 \pm \sqrt 5) is correct because these are the equations of the latus rectum of the ellipse after rewriting it in standard form and calculating the eccentricity.
68 Medium
68 of 187
Application +4 | -1
The equation of ellipse, whose distance between the foci is equal to 8 and the distance between the directrices is 18 , is
View answer and explanation
Correct option: D
5x2+9y2=1805 x^2+9 y^2=180
Option D (5x2+9y2=1805 x^2+9 y^2=180) is correct because it satisfies the given conditions for the distance between foci and directrices.
69 Medium
69 of 187
Application +4 | -1
An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the necessary length of the string and the distance between the pins respectively in cms. Are
View answer and explanation
Correct option: D
6+25,256 + 2\sqrt 5 ,2\sqrt 5
Option D (6+25,256 + 2\sqrt 5 ,2\sqrt 5) is correct because the length of the string is the sum of the major axis and the distance between the foci, and the distance between the pins is the distance between the foci.
70 Medium
70 of 187
Application +4 | -1
Let x2a2+y2b2=1(a>b)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) be a given ellipse, length of whose latusrectum is 10 . If its eccentricity is the maximum value of the function ϕ(t)=512+t−t2\phi(t)=\frac{5}{12}+t-t^2, then a2+b2a^2+b^2 is equal to
View answer and explanation
Correct option: B
126
Option B (126) is correct because the values of a and b are calculated using the latus rectum and eccentricity, and then a2+b2a^2 + b^2 is computed.
71 Medium
71 of 187
Understanding +4 | -1
The product of the perpendiculars from the foci of the ellipse x2144+y2100=1\frac{x^2}{144}+\frac{y^2}{100}=1 on any tangent is:
View answer and explanation
Correct option: B
100
Option B (100) is correct because the product of the perpendiculars from the foci of an ellipse to any tangent is equal to the square of the semi-minor axis.
72 Medium
72 of 187
Analyzing +4 | -1
If a triangle is inscribed in an ellipse and two of its sides are parallel to given straight lines, then envelope of the third side is
View answer and explanation
Correct option: D
An ellipse
Option D (An ellipse) is correct because the envelope of the third side of the triangle is an ellipse when two sides are parallel to given straight lines.
73 Medium
73 of 187
Application +4 | -1
Let E\mathrm{E} be the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 and CC be the circle x2+y2=9x^2+y^2=9. Let PP and QQ be the points (1,2)(1,2) and (2,1)(2,1), respectively. Then
View answer and explanation
Correct option: D
PP lies inside CC but outside EE
Option D (P lies inside C but outside E) is correct because substituting the coordinates of P and Q into the equations of the ellipse and circle determines their positions relative to the curves.
74 Medium
74 of 187
Application +4 | -1
The tangent at the point P(42,32)P\left(\frac{4}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) to the ellipse 9x2+16y2=1449 x^2+16 y^2=144 meets the axis of xx at AA and the axis of yy at BB. If CC is the centre of the ellipse, then area of the ΔABC\Delta A B C is (in sq. units)
View answer and explanation
Correct option: A
12
Option B (12) is correct because it is the area of the triangle formed by the tangent to the ellipse and the coordinate axes.
75 Medium
75 of 187
Application +4 | -1
The locus of points whose polars with respect to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 are at a distance dd from the centre of the ellipse, is
View answer and explanation
Correct option: B
x2a4+y2b4=1d2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{d^2}}}
Option B (x2a4+y2b4=1d2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{d^2}}}) is correct because it represents the locus of points whose polars are at a distance d from the center of the ellipse.
76 Medium
76 of 187
Application +4 | -1
If roots of quadratic equation ax2+2bx+c=0a{x^2} + 2bx + c = 0 are not real, then ax2+2bxy+cy2+dx+ey+f=0a{x^2} + 2bxy + c{y^2} + dx + ey + f = 0 represents a/an
View answer and explanation
Correct option: A
Ellipse
Option A (Ellipse) is correct because if the roots of the quadratic equation are not real, then b2<acb^2 < ac, which implies the given equation represents an ellipse.
77 Medium
77 of 187
Application +4 | -1
The eccentricity of the ellipse 25x2+16y2=40025 x^2+16 y^2=400 is
View answer and explanation
Correct option: A
3/53 / 5
Option A (3/53 / 5) is correct because the eccentricity is calculated using the formula e=1−a2b2e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}} since b>ab > a.
78 Medium
78 of 187
Application +4 | -1
Let PiP_i and Pi′P_i^{\prime} be the feet of the perpendiculars drawn from the foci SS and S′S^{\prime} on a tangent TiT_i to an ellipse whose length of semi-major axis is 20 . If ∑i=110(SPi)(S′Pi′)=2560\sum\limits_{{i=1}}^{10}\left(S P_i\right)\left(S^{\prime} P_i^{\prime}\right)=2560, then the value of eccentricity is
View answer and explanation
Correct option: C
35\frac{3}{5}
Option C (35\frac{3}{5}) is correct because the eccentricity is calculated using the relationship between the semi-major axis, semi-minor axis, and eccentricity of the ellipse.
79 Medium
79 of 187
Application +4 | -1
Chords of an ellipse are drawn through the positive end of the minor axis. Then, their mid-point lies on
View answer and explanation
Correct option: C
an ellipse
Option C (an ellipse) is correct because the locus of the mid-point of the chords is found to be an ellipse by using the equation of the chord and the condition that it passes through the end of the minor axis.
80 Medium
80 of 187
Application +4 | -1
If the tangents on the ellipse 4x2+y2=84{x^2} + {y^2} = 8 at the points (1, 2) and (a, b)(a,\,b) are perpendicular to each other, then a2{a^2} is equal to
View answer and explanation
Correct option: D
217\frac{2}{{17}}
Option D (217\frac{{2}}{{17}}) is correct because it is the value of a2{a^2} obtained by using the perpendicularity condition of tangents and the equation of the ellipse.
81 Medium
81 of 187
Analyzing +4 | -1
The length of the major axis of the ellipse (5x−10)2+(5y+15)2=(3x−4y+7)24(5 x-10)^2+(5 y+15)^2=\frac{(3 x-4 y+7)^2}{4} is
View answer and explanation
Correct option: B
203\frac{20}{3}
Option B (203\frac{20}{3}) is correct because it is calculated by finding the distance from the focus to the directrix and using the eccentricity to find the length of the major axis.
82 Medium
82 of 187
Analyzing +4 | -1
If the point of intersection of the ellipses x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and x2α2+y2β2=1\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1 are at the extremities of the conjugate diameters of the former, then
View answer and explanation
Correct option: A
a2α2+b2β2=2\frac{a^2}{\alpha^2}+\frac{b^2}{\beta^2}=2
Option A (a2α2+b2β2=2\frac{a^2}{\alpha^2}+\frac{b^2}{\beta^2}=2) is correct because it represents the relationship between the semi-major and semi-minor axes of the two ellipses when their intersection points are at the extremities of conjugate diameters.
83 Medium
83 of 187
Application +4 | -1
The minimum distance of the centre of the ellipse x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1 from the chord of contact of mutually perpendicular tangents of the ellipse is
View answer and explanation
Correct option: B
165\frac{16}{5}
Option B (165\frac{16}{5}) is correct because it represents the minimum distance from the center of the ellipse to the chord of contact of mutually perpendicular tangents, which is related to the radius of the director circle.
84 Medium
84 of 187
Analyzing +4 | -1
If the chord of contact of tangents drawn from a point PP to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 subtends a right-angle at its centre, then PP lies on
View answer and explanation
Correct option: D
x2a4+y2b4=1a2+1b2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}
Option D (x2a4+y2b4=1a2+1b2\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}) is correct because it represents the locus of point P such that the chord of contact subtends a right angle at the center of the ellipse.
85 Medium
85 of 187
Application +4 | -1
A man running round a race course notes that the sum of the distances of two flag posts from him is always 10m and the distance between the flag posts is 8m. The area of the path he encloses (in square metre) is
View answer and explanation
Correct option: A
15π15\pi
Option A (15π15\pi) is correct because the area of the elliptical path is calculated using the formula A=πabA = \pi ab, where a and b are the semi-major and semi-minor axes, respectively.
86 Medium
86 of 187
Analyzing +4 | -1
Tangent at a point on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is drawn which cuts the coordinates axes at AA and BB. The minimum area of the triangle OABO A B is ( OO being origin)
View answer and explanation
Correct option: A
aba b
Option D ((4am2l2,4aml)\left(\frac{4 a m^2}{l^2}, \frac{4 a m}{l}\right)) is correct because it represents the coordinates of point D, which is the intersection of the third normal from point C with the parabola.
87 Medium
87 of 187
Application +4 | -1
The foci of the ellipse x216+y2b2=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1 and the hyperbola x2144−y281=125\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}} coincide. Then, the value of b2{b^2} is
View answer and explanation
Correct option: C
7
Option C (7) is correct because the value of b² is calculated by equating the foci of the ellipse and hyperbola.
88 Medium
88 of 187
Analyzing +4 | -1
If the normal at the end of latusrectum of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 passes through (0,  −b)\left( {0,\; - b} \right), then e4+e2{e^4} + {e^2} (where ee is eccentricity) equals
View answer and explanation
Correct option: A
1
The correct answer is 1 because the given condition implies a relationship between the eccentricity and the semi-major/minor axes, leading to e4+e2=1{e^4} + {e^2} = 1.
89 Medium
89 of 187
Application +4 | -1
Let (α,β)\left( {\mathrm{\alpha },\mathrm{\beta }} \right) be a point from which two perpendicular tangents can be drawn to the ellipse 4x2+5y2=20.4{x^2} + 5{y^2} = 20. If F=4α+3βF = 4\mathrm{\alpha } + 3\mathrm{\beta }, then
View answer and explanation
Correct option: A
−15≤F≤15- 15 \le F \le 15
Option A (−15≤F≤15- 15 \le F \le 15) is correct because the point (α,β)(\alpha, \beta) lies on the director circle of the ellipse, and by expressing α\alpha and β\beta in terms of trigonometric functions, we can find the range of FF.
90 Medium
90 of 187
Analyzing +4 | -1
PP is a point on the circle x2+y2=c2.{x^2} + {y^2} = {c^2}. The locus of the mid-points of chords of contact of PP with respect to x2a2+y2b2=1,\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, is
View answer and explanation
Correct option: A
c2(x2a2+y2b2)=x2+y2{c^2}\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right) = {x^2} + {y^2}
Option A (c2(x2a2+y2b2)=x2+y2{c^2}(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}) = {x^2} + {y^2}) is correct because it represents the locus of the midpoints of the chords of contact of tangents from a point on the circle to the ellipse.
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