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MCQMediumJEE 2025Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2025 Question with Solution

Let the ellipse 3x2+py2=43x^2 + py^2 = 4 pass through the centre CC of the circle x2+y2−2x−4y−11=0x^2 + y^2 - 2x - 4y - 11 = 0 of radius rr. Let f1,f2f_1, f_2 be the focal distances of the point CC on the ellipse. Then 6f1f2−r6f_1 f_2 - r is equal to

  • A

    7070

  • B

    6868

  • C

    7878

  • D

    7474

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The ellipse is 3x2+py2=43x^2 + py^2 = 4 and it passes through the centre CC of the circle x2+y2−2x−4y−11=0x^2 + y^2 - 2x - 4y - 11 = 0.

Find: The value of 6f1f2−r6f_1 f_2 - r.

First, rewrite the circle by completing the square:

(x−1)2+(y−2)2=16(x-1)^2 + (y-2)^2 = 16

Hence, the centre is C=(1,2)C = (1,2) and the radius is r=4r = 4.

Since the ellipse passes through CC, substitute x=1,y=2x=1, y=2 into 3x2+py2=43x^2 + py^2 = 4:

3(1)2+p(2)2=43(1)^2 + p(2)^2 = 4 3+4p=43 + 4p = 4 p=14p = \frac{1}{4}

So the ellipse becomes

3x2+14y2=43x^2 + \frac{1}{4}y^2 = 4

which can be written as

x243+y216=1\frac{x^2}{\frac{4}{3}} + \frac{y^2}{16} = 1

Therefore, the semi-axes are a2=43a^2 = \frac{4}{3} and b2=16b^2 = 16, with major axis along the yy-axis.

The focal distance satisfies

c2=b2−a2=16−43=443c^2 = b^2 - a^2 = 16 - \frac{4}{3} = \frac{44}{3}

so

c=2333c = \frac{2\sqrt{33}}{3}

Using the focal distance expressions for the point C=(1,2)C=(1,2) on this ellipse:

f1=4+333,f2=4−333f_1 = 4 + \frac{\sqrt{33}}{3}, \qquad f_2 = 4 - \frac{\sqrt{33}}{3}

Therefore,

f1f2=(4+333)(4−333)f_1f_2 = \left(4 + \frac{\sqrt{33}}{3}\right)\left(4 - \frac{\sqrt{33}}{3}\right) =16−339=16−113=373= 16 - \frac{33}{9} = 16 - \frac{11}{3} = \frac{37}{3}

Now compute the required value:

6f1f2−r=6⋅373−4=74−4=706f_1f_2 - r = 6 \cdot \frac{37}{3} - 4 = 74 - 4 = 70

Therefore, the correct option is A.

Using distances from the two foci

Given: C=(1,2)C=(1,2) lies on the ellipse, and the circle has radius rr.

Find: 6f1f2−r6f_1f_2-r.

From the circle,

(x−1)2+(y−2)2=16(x-1)^2 + (y-2)^2 = 16

so r=4r=4.

Substituting C=(1,2)C=(1,2) in the ellipse gives

3+4p=4⇒p=143 + 4p = 4 \Rightarrow p=\frac{1}{4}

Hence,

x243+y216=1\frac{x^2}{\frac{4}{3}} + \frac{y^2}{16} = 1

The foci are

(0,±2333)\left(0, \pm \frac{2\sqrt{33}}{3}\right)

Now the two focal distances are

f1=12+(2−2333)2,f2=12+(2+2333)2f_1 = \sqrt{1^2 + \left(2-\frac{2\sqrt{33}}{3}\right)^2}, \qquad f_2 = \sqrt{1^2 + \left(2+\frac{2\sqrt{33}}{3}\right)^2}

The extracted solution simplifies these to

f1=4+333,f2=4−333f_1 = 4 + \frac{\sqrt{33}}{3}, \qquad f_2 = 4 - \frac{\sqrt{33}}{3}

and hence

f_1f_2 = \frac{37}{3} $$.

Therefore,

6f1f2−r=6⋅373−4=706f_1f_2-r = 6\cdot \frac{37}{3} - 4 = 70

So the correct option is A.

Common mistakes

  • A common mistake is completing the square incorrectly for the circle and getting the centre wrong. If CC is not found as (1,2)(1,2), then both pp and the final value become incorrect. Always rewrite the circle first in standard form.

  • Students often identify the wrong major axis of the ellipse. After converting to standard form, compare 43\frac{4}{3} and 1616 carefully. Since 1616 is larger, the major axis is along the yy-axis, not the xx-axis.

  • Another mistake is using an incorrect property such as taking the product f1f2f_1f_2 to be constant for every point on the ellipse. The reliable method here is to use the expressions obtained in the solution and then multiply them carefully.

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