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Ellipse question

2025 · 24 Jan · Shift 1 · Q33
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  5. /2025 · 24 Jan · Shift 1 · Q33

Ellipse question

2025 · 24 Jan · Shift 1 · Q33

JEE MainMathematicsEllipseMCQ+4 / −1
Let the product of the focal distances of the point (3,12)\left(\sqrt{3}, \frac{1}{2}\right)(3​,21​) on the ellipse x2a2+y2b2=1,(a>b)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)a2x2​+b2y2​=1,(a>b), be 74\frac{7}{4}47​. Then the absolute difference of the eccentricities of two such ellipses is
  1. A
    1−223\frac{1-2 \sqrt{2}}{\sqrt{3}}3​1−22​​
  2. B
    1−32\frac{1-\sqrt{3}}{\sqrt{2}}2​1−3​​
  3. C
    3−2223\frac{3-2 \sqrt{2}}{2 \sqrt{3}}23​3−22​​
  4. D
    3−2232\frac{3-2 \sqrt{2}}{3 \sqrt{2}}32​3−22​​
View written solutionFree

Correct answer: C

  1. Set up the ellipse and the given point

The ellipse is

x2a2+y2b2=1,a>b.\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b.a2x2​+b2y2​=1,a>b.

Its eccentricity is

e=ca,c2=a2−b2.e=\frac{c}{a}, \qquad c^2=a^2-b^2.e=ac​,c2=a2−b2.

The point

P(3,12)P\left(\sqrt{3},\frac12\right)P(3​,21​)

lies on the ellipse, so

3a2+1/4b2=1.\frac{3}{a^2}+\frac{1/4}{b^2}=1.a23​+b21/4​=1.

Equivalently,

3a2+14b2=1.(1)\frac{3}{a^2}+\frac{1}{4b^2}=1. \tag{1}a23​+4b21​=1.(1)
  1. Use the product of focal distances

For a point (x,y)(x,y)(x,y) on the ellipse, if r1,r2r_1,r_2r1​,r2​ are its distances from the two foci, then

r1r2=b2+e2x2.r_1r_2=b^2+e^2x^2.r1​r2​=b2+e2x2.

Here x=3x=\sqrt{3}x=3​ and it is given that

r1r2=74.r_1r_2=\frac74.r1​r2​=47​.

Hence,

b2+3e2=74.b^2+3e^2=\frac74.b2+3e2=47​.

Since

e2=1−b2a2,e^2=1-\frac{b^2}{a^2},e2=1−a2b2​,

we get

b2+3(1−b2a2)=74.b^2+3\left(1-\frac{b^2}{a^2}\right)=\frac74.b2+3(1−a2b2​)=47​.

So,

b2+3−3b2a2=74b^2+3-\frac{3b^2}{a^2}=\frac74b2+3−a23b2​=47​ b2−3b2a2=−54b^2-\frac{3b^2}{a^2}=-\frac54b2−a23b2​=−45​ b2(1−3a2)=−54.(2)b^2\left(1-\frac{3}{a^2}\right)=-\frac54. \tag{2}b2(1−a23​)=−45​.(2)
  1. Use the point-on-ellipse condition

From (1):

3a2+14b2=1.\frac{3}{a^2}+\frac{1}{4b^2}=1.a23​+4b21​=1.

Multiply by 4b24b^24b2:

12b2a2+1=4b2\frac{12b^2}{a^2}+1=4b^2a212b2​+1=4b2 4b2−12b2a2=14b^2-\frac{12b^2}{a^2}=14b2−a212b2​=1 4b2(1−3a2)=1.(3)4b^2\left(1-\frac{3}{a^2}\right)=1. \tag{3}4b2(1−a23​)=1.(3)

From (2):

b2(1−3a2)=−54,b^2\left(1-\frac{3}{a^2}\right)=-\frac54,b2(1−a23​)=−45​,

which seems inconsistent with (3). So let us re-check the focal-distance product formula carefully.

  1. Correct focal-distance product formula

For ellipse with foci (±c,0)(\pm c,0)(±c,0), for a point (x,y)(x,y)(x,y) on it,

r12=(x−c)2+y2,r22=(x+c)2+y2.r_1^2=(x-c)^2+y^2, \qquad r_2^2=(x+c)^2+y^2.r12​=(x−c)2+y2,r22​=(x+c)2+y2.

Thus,

(r1r2)2=((x−c)2+y2)((x+c)2+y2)(r_1r_2)^2=\big((x-c)^2+y^2\big)\big((x+c)^2+y^2\big)(r1​r2​)2=((x−c)2+y2)((x+c)2+y2) =(x2+y2+c2)2−4c2x2.= (x^2+y^2+c^2)^2-4c^2x^2.=(x2+y2+c2)2−4c2x2.

Using the ellipse relation, this simplifies to

r1r2=a2e2y2b2+b2.r_1r_2=a^2e^2\frac{y^2}{b^2}+b^2.r1​r2​=a2e2b2y2​+b2.

Since c2=a2e2c^2=a^2e^2c2=a2e2 and also a2e2=a2−b2a^2e^2=a^2-b^2a2e2=a2−b2, a standard usable form is

r1r2=b2+c2y2b2.r_1r_2=b^2+\frac{c^2y^2}{b^2}.r1​r2​=b2+b2c2y2​.

Here y=12y=\frac12y=21​, so

r1r2=b2+c24b2=74.r_1r_2=b^2+\frac{c^2}{4b^2}=\frac74.r1​r2​=b2+4b2c2​=47​.

Therefore,

b2+a2−b24b2=74.(4)b^2+\frac{a^2-b^2}{4b^2}=\frac74. \tag{4}b2+4b2a2−b2​=47​.(4)
  1. Convert to simpler variables

Let

A=a2,B=b2.A=a^2, \qquad B=b^2.A=a2,B=b2.

Then from the point condition,

3A+14B=1.(5)\frac{3}{A}+\frac{1}{4B}=1. \tag{5}A3​+4B1​=1.(5)

And from (4),

B+A−B4B=74.(6)B+\frac{A-B}{4B}=\frac74. \tag{6}B+4BA−B​=47​.(6)

From (6), multiply by 4B4B4B:

4B2+A−B=7B4B^2+A-B=7B4B2+A−B=7B A+4B2−8B=0A+4B^2-8B=0A+4B2−8B=0 A=8B−4B2.(7)A=8B-4B^2. \tag{7}A=8B−4B2.(7)

Substitute into (5):

38B−4B2+14B=1.\frac{3}{8B-4B^2}+\frac{1}{4B}=1.8B−4B23​+4B1​=1.

Factor:

34B(2−B)+14B=1.\frac{3}{4B(2-B)}+\frac{1}{4B}=1.4B(2−B)3​+4B1​=1.

Multiply by 4B(2−B)4B(2-B)4B(2−B):

3+(2−B)=4B(2−B)3+(2-B)=4B(2-B)3+(2−B)=4B(2−B) 5−B=8B−4B25-B=8B-4B^25−B=8B−4B2 4B2−9B+5=0.4B^2-9B+5=0.4B2−9B+5=0.

So,

(4B−5)(B−1)=0.(4B-5)(B-1)=0.(4B−5)(B−1)=0.

Hence,

B=1orB=54.B=1 \quad \text{or} \quad B=\frac54.B=1orB=45​.

Using (7):

  • If B=1B=1B=1, then A=8−4=4.A=8-4=4.A=8−4=4.
  • If B=54B=\frac54B=45​, then A=8⋅54−4⋅2516=10−254=154.A=8\cdot\frac54-4\cdot\frac{25}{16}=10-\frac{25}{4}=\frac{15}{4}.A=8⋅45​−4⋅1625​=10−425​=415​.

So the two ellipses are:

(a2,b2)=(4,1),(154,54).(a^2,b^2)=(4,1), \qquad \left(\frac{15}{4},\frac54\right).(a2,b2)=(4,1),(415​,45​).
  1. Find their eccentricities

For the first ellipse:

e1=1−b2a2=1−14=32.e_1=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac14}=\frac{\sqrt3}{2}.e1​=1−a2b2​​=1−41​​=23​​.

For the second ellipse:

e2=1−54154=1−13=23=63.e_2=\sqrt{1-\frac{\frac54}{\frac{15}{4}}} =\sqrt{1-\frac13} =\sqrt{\frac23} =\frac{\sqrt6}{3}.e2​=1−415​45​​​=1−31​​=32​​=36​​.
  1. Compute the absolute difference
∣e1−e2∣=∣32−63∣.|e_1-e_2|=\left|\frac{\sqrt3}{2}-\frac{\sqrt6}{3}\right|.∣e1​−e2​∣=​23​​−36​​​.

Taking positive difference,

∣e1−e2∣=32−63=33−266.|e_1-e_2|=\frac{\sqrt3}{2}-\frac{\sqrt6}{3} =\frac{3\sqrt3-2\sqrt6}{6}.∣e1​−e2​∣=23​​−36​​=633​−26​​.

Factor 3\sqrt33​:

=3(3−22)6=3−2223.=\frac{\sqrt3(3-2\sqrt2)}{6} =\frac{3-2\sqrt2}{2\sqrt3}.=63​(3−22​)​=23​3−22​​.
  1. Match with the options

This is exactly Option C:

3−2223.\boxed{\frac{3-2\sqrt2}{2\sqrt3}}.23​3−22​​​.
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