$y=m x+c$ is a normal to the ellipse $x^2 / a^2+y^2 / b^2=1$ if $c^2$ is equal to

$\frac{\left(a^2-b^2\right)^2}{a^2 m^2+b^2}$

$\frac{\left(a^2-b^2\right)^2}{a^2 m^2}$

Tangents are drawn from the point on the line $x - y - 5 = 0$ to ${x^2} + 4{y^2} = 4$, then all the chords of contact pass through a fixed point, whose coordinates are

$\left( {\frac{1}{5},\;\frac{4}{5}} \right)$

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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Medium

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Showing 151-180 of 187 questions

151 Medium

151 of 187

Application +4 | -1

If the normal at any point

P

on the ellipse

\frac{x^2}{64}+\frac{y^2}{36}=1

meets the major axis at

G_1

and the minor axis at

G_2

then the ratio of

P G_1

and

P G_2

is equal to

View answer and explanation

Correct option: C

9: 16

Option D (

16: 9

) is incorrect. Option C (

9: 16

) is correct because it represents the ratio of

P G_1

and

P G_2

. The correct ratio is derived from the geometry of the ellipse and its normal.

152 Medium

152 of 187

Application +4 | -1

The locus of the foot of the perpendicular drawn from the centre on any tangent to the ellipse

\frac{x^2}{25}+\frac{y^2}{16}=1

is:

View answer and explanation

Correct option: A

25x^2 + 16y^2 = (x^2 + y^2)^2

Option A (

25x^2 + 16y^2 = (x^2 + y^2)^2

) is correct because it represents the locus of the foot of the perpendicular from the center to any tangent of the ellipse.

153 Medium

153 of 187

Analyzing +4 | -1

Consider the ellipse

\frac{{{x^2}}}{{f\left( {{K^2} + 2K + 5} \right)}} + \frac{{{y^2}}}{{f\left( {K + 11} \right)}} = 1,

where

f(x)

is a positive decreasing function, then the value (s) of

K

for which the major axis coincide with x – axis, is

View answer and explanation

Correct option: C

K \in \left( { - 3,2} \right)

Option C (

K \in ( - 3,2)

) is correct because for the major axis to coincide with the x-axis, the coefficient of x² must be greater than the coefficient of y², leading to the inequality

K^2 + 2K + 5 < K + 11

154 Medium

154 of 187

Application +4 | -1

The area of the rectangle formed by the perpendiculars from the centre of the ellipse to the tangent and normal at the point-whose eccentric angle is

\pi / 4

, is

View answer and explanation

Correct option: A

\left(\frac{a^2-b^2}{a^2+b^2}\right) a b

Option A (

\left(\frac{a^2-b^2}{a^2+b^2}\right) a b

) is correct because the area of the rectangle is the product of the perpendicular distances from the center to the tangent and normal, which simplifies to the given expression.

155 Medium

155 of 187

Application +4 | -1

y=m x+c

is a normal to the ellipse

x^2 / a^2+y^2 / b^2=1

c^2

is equal to

View answer and explanation

Correct option: C

\frac{\left(a^2-b^2\right)^2 m^2}{a^2+b^2 m^2}

Option D (

\frac{\left(a^2-b^2\right)^2 m^2}{a^2 m^2+b^2}

) is correct because it provides the correct condition for the line to be a normal to the ellipse.

156 Medium

156 of 187

Analyzing +4 | -1

If the length of latusrectum of ellipse

E_1: 4(x+y-1)^2+2(x-y+3)^2=8

and

E_2: \frac{x^2}{p}+\frac{y^2}{p^2}=1,(0 < p < 1)

are equal, then area of ellipse

E_2

, is

View answer and explanation

Correct option: B

\frac{\pi}{\sqrt{2}}

Option B (

\frac{{\pi}}{{\sqrt{{2}}}}

) is correct because after finding the length of the latus rectum of ellipse E1 and equating it to the length of the latus rectum of ellipse E2, we can find p and then the area of ellipse E2.

157 Medium

157 of 187

Application +4 | -1

If the area of the triangle formed by the three points on an ellipse, whose eccentric angle are

\theta, \phi

, and

\psi

, is Kabsin

\frac{\varphi-\psi}{2} \sin \frac{\psi-\theta}{2} \sin \frac{\theta-\varphi}{2}

, then the value of

K

View answer and explanation

Correct option: A

Option A (2) is correct because the area of the triangle formed by the points on the ellipse is Kabsin((φ-ψ)/2)sin((ψ-θ)/2)sin((θ-φ)/2), where K equals 2.

158 Medium

158 of 187

Remembering +4 | -1

The sum of focal distances of any point on the ellipse with major and minor axes,

2 a

and

2 b

, respectively, is equal to

View answer and explanation

Correct option: A

2 a

Option A (

2 a

) 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

159 Medium

159 of 187

Application +4 | -1

If the distance between the focii of an ellipse is equal to the length of the lectus rectum, the eccentricity of the ellipse is :

View answer and explanation

Correct option: B

\frac{\sqrt{5}-1}{2}

Option D (-12) is correct because by comparing the given normal equation to the standard form, the value of k is found to be -12.

160 Medium

160 of 187

Application +4 | -1

An ellipse slides between two perpendicular lines. The locus of its centre, is

View answer and explanation

Correct option: A

a circle

Option A (a circle) is correct because the locus of the center of an ellipse sliding between two perpendicular lines is a circle.

161 Medium

161 of 187

Analyzing +4 | -1

The normal at a variable point

P

on an ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

of eccentricity e meets the axes of the ellipse in

Q

and

R

, then the locus of the mid-point of

Q R

is a conic with eccentricity

e^{\prime}

such that

View answer and explanation

Correct option: C

e^{\prime}=e

Option C (

e^{\prime}=e

) is correct because the locus of the midpoint of QR is an ellipse with the same eccentricity as the original ellipse.

162 Medium

162 of 187

Application +4 | -1

The point on the ellipse

4 x^2+9 y^2=1

where the tangent drawn is parallel to line

8 x=9 y

View answer and explanation

Correct option: B

\left(\frac{-2}{5}, \frac{1}{5}\right)

Option B (

\left(\frac{{-2}}{{5}}, \frac{{1}}{{5}}\right)

) is correct because it represents the point on the ellipse where the tangent is parallel to the given line.

163 Medium

163 of 187

Analyzing +4 | -1

(a \cos \alpha, b \sin \alpha),(a \cos \beta, b \sin \beta)

are the end points of a focal chord, then which of the following is correct?

View answer and explanation

Correct option: C

\frac{e-1}{e+1}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2}

Option C (

\frac{e-1}{e+1}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2}

) is correct because it represents the relationship between the eccentric angles of the endpoints of a focal chord and the eccentricity of the ellipse.

164 Medium

164 of 187

Application +4 | -1

P(x, y), F_1(3,0), F_2(-3,0)

and

16 x^2+25 y^2=400

, then

P F_1+P F_2

equals

View answer and explanation

Correct option: C

Option C (10) is correct because

PF_1 + PF_2 = 2a

for any point P on the ellipse, and

a=5

in this case.

165 Medium

165 of 187

Analyzing +4 | -1

Locus of all such points from where the tangents drawn to the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

are always inclined at

45^{\circ}

, is

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the correct locus equation differs from the provided options. The correct equation is derived using the condition for the angle between tangents.

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166 Medium

166 of 187

Analyzing +4 | -1

The locus of the mid-points of the chords of the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

whose poles are on the auxiliary circle on the tangents at the extremities of which intersect on the auxiliary circle, is

View answer and explanation

Correct option: D

x^2+y^2=a^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2

Option D (

x^2+y^2=a^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2

) is correct because it represents the locus of mid-points of chords whose poles lie on the auxiliary circle.

167 Medium

167 of 187

Application +4 | -1

The equation of the ellipse whose axes are coincident with the coordinate axes and which touches the straight line

3x-2y-20 = 0

and

x+6 y-20=0

, is

View answer and explanation

Correct option: C

\frac{x^2}{40}+\frac{y^2}{10}=1

Option C (

\frac{x^2}{40}+ \frac{y^2}{10}=1

) is correct because it represents the equation of the ellipse that touches both given lines, found by using the tangent condition.

168 Medium

168 of 187

Analyzing +4 | -1

The point, at shortest distance from the line

x + y = 7

and lying on an ellipse

{x^2} + 2{y^2} = 6

, has coordinates

View answer and explanation

Correct option: C

\left( {2,1} \right)

Option C ((2,1)) is correct because the tangent to the ellipse at this point is parallel to the given line x + y = 7.

169 Medium

169 of 187

Application +4 | -1

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is

\frac{3}{2}

units, then its eccentricity is?

View answer and explanation

Correct option: D

\frac{1}{3}

Option D (

\frac{{1}}{{3}}

) is correct because it is derived from the given latus rectum length and the distance between a focus and its nearest vertex.

170 Medium

170 of 187

Application +4 | -1

The sum of the squares of the perpendiculars on any tangent to the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

from two points on the minor axis each at a distance

\sqrt {{a^2} - {b^2}}

from the centre is

View answer and explanation

Correct option: A

2{a^2}

Option A (

2{a^2}

) is correct because the sum of the squares of the perpendiculars from the specified points on the minor axis to any tangent of the ellipse simplifies to

2a^2

171 Medium

171 of 187

Application +4 | -1

The eccentricity of the ellipse, whose latus rectum is equal to the distance between two focus points, is

View answer and explanation

Correct option: B

\frac{\sqrt{5}-1}{2}

Option B (

\frac{{\sqrt{{5}}-1}}{{2}}

) is correct because it is the positive solution to the equation derived from the given condition relating latus rectum and focal distance.

172 Medium

172 of 187

Application +4 | -1

p

is the length of the perpendicular from a facus upon the tangent at any point

p

of the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

and

r

is the distance of

p

from the focus, then

\frac{{2a}}{r} - \frac{{{b^2}}}{{{p^2}}}

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because, after substituting and simplifying, the expression equals 1.

173 Medium

173 of 187

Application +4 | -1

Let

E

be the ellipse

\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1

and

C

be the circle

{x^2} + {y^2} = 9

. Let

P

and

Q

be the points

\left( {1,2} \right)

and

\left( {2,1} \right)

respectively. Then

View answer and explanation

Correct option: D

P

lies inside

C

but outside

E

Option D (

P

lies inside

C

but outside

E

) is correct because substituting the coordinates of P into the equations of the circle and ellipse shows that P is inside the circle and outside the ellipse.

174 Medium

174 of 187

Understanding +4 | -1

The equation of a diameter conjugate to a diameter

y = \frac{b}{a}x

of the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

View answer and explanation

Correct option: A

y = - \frac{b}{a}x

Option A (

y = - \frac{b}{a}x

) is correct because conjugate diameters of an ellipse satisfy the condition

{m_1}{m_2} = - \frac{{{b^2}}}{{{a^2}}}

175 Medium

175 of 187

Application +4 | -1

The locus of the point of intersection of tangents to the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

which meet at right angle is

View answer and explanation

Correct option: A

A circle

Option A A circle is correct because the locus of the point of intersection of perpendicular tangents to an ellipse is its director circle, which is a circle.

176 Medium

176 of 187

Understanding +4 | -1

In an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on

View answer and explanation

Correct option: B

Corresponding directrix

Option B (Corresponding directrix) is correct because the intersection point lies on the directrix of the ellipse, as derived from the tangent and perpendicular equations.

177 Medium

177 of 187

Application +4 | -1

C

is the centre of the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

and the normal at an end of a latusrectum cuts the major axis in

G,

then

CG =

View answer and explanation

Correct option: C

a{e^2}

Option C is correct because the distance CG is calculated using the equation of the normal at the end of the latus rectum and the eccentricity of the ellipse.

178 Medium

178 of 187

Application +4 | -1

The locus of the middle points of the portions of the tangents of the ellipse

x^2 / a^2+y^2 / b^2=1

included between the axis is the curve.

View answer and explanation

Correct option: B

a^2 / x^2+b^2 / y^2=4

Option B (

a^2 / x^2+b^2 / y^2=4

) is correct because it represents the locus of the midpoints of the tangent segments of the ellipse intercepted by the axes.

179 Medium

179 of 187

Application +4 | -1

The line

x = a{t^2}

meets the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

in the real points, if

View answer and explanation

Correct option: B

\left| t \right| \le 1

Option B (

\left| t \right| \le 1

) is correct because it ensures that the value inside the square root for calculating y is non-negative, resulting in real points of intersection.

180 Medium

180 of 187

Application +4 | -1

Tangents are drawn from the point on the line

x - y - 5 = 0

{x^2} + 4{y^2} = 4

, then all the chords of contact pass through a fixed point, whose coordinates are

View answer and explanation

Correct option: D

\left( {\frac{4}{5},\; - \frac{1}{5}} \right)

Option D (

\left( {\frac{4}{5},\; - \frac{1}{5}} \right)

) is correct because the chord of contact passes through a fixed point which is the intersection of two lines derived from the equation of the chord of contact.

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