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Ellipse question

2025 · 4 Apr · Shift 2 · Q35
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  5. /2025 · 4 Apr · Shift 2 · Q35

Ellipse question

2025 · 4 Apr · Shift 2 · Q35

JEE MainMathematicsEllipseMCQ+4 / −1
The centre of a circle C is at the centre of the ellipse E:x2a2+y2 b2=1,a>b\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}\gt \mathrm{b}E:a2x2​+ b2y2​=1,a>b. Let C pass through the foci F1F_1F1​ and F2F_2F2​ of E such that the circle CCC and the ellipse EEE intersect at four points. Let P be one of these four points. If the area of the triangle PF1 F2\mathrm{PF}_1 \mathrm{~F}_2PF1​ F2​ is 30 and the length of the major axis of EEE is 17 , then the distance between the foci of EEE is :
  1. A
    12
  2. B
    26
  3. C
    13
  4. D
    132\frac{13}{2}213​
View written solutionFree

Correct answer: C

  1. Set up the ellipse and circle

The ellipse is

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

Its centre is the origin, and its foci are

F1(−c,0),F2(c,0),c2=a2−b2.F_1(-c,0),\quad F_2(c,0),\qquad c^2=a^2-b^2.F1​(−c,0),F2​(c,0),c2=a2−b2.

The circle CCC is centered at the centre of the ellipse and passes through the foci, so its radius is ccc. Hence its equation is

x2+y2=c2.x^2+y^2=c^2.x2+y2=c2.
  1. Find the intersection points of the circle and ellipse

At any intersection point P(x,y)P(x,y)P(x,y), both equations hold:

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

and

x2+y2=c2.x^2+y^2=c^2.x2+y2=c2.

Using c2=a2−b2c^2=a^2-b^2c2=a2−b2, from the circle we get

y2=c2−x2.y^2=c^2-x^2.y2=c2−x2.

Substitute into the ellipse:

x2a2+c2−x2b2=1.\frac{x^2}{a^2}+\frac{c^2-x^2}{b^2}=1.a2x2​+b2c2−x2​=1.

Multiply by a2b2a^2b^2a2b2:

b2x2+a2c2−a2x2=a2b2.b^2x^2+a^2c^2-a^2x^2=a^2b^2.b2x2+a2c2−a2x2=a2b2.

So,

(b2−a2)x2=a2(b2−c2).(b^2-a^2)x^2=a^2(b^2-c^2).(b2−a2)x2=a2(b2−c2).

Now use c2=a2−b2c^2=a^2-b^2c2=a2−b2:

b2−c2=b2−(a2−b2)=2b2−a2.b^2-c^2=b^2-(a^2-b^2)=2b^2-a^2.b2−c2=b2−(a2−b2)=2b2−a2.

But it is simpler to proceed directly:

(b2−a2)x2=a2b2−a2c2=a2(b2−c2).(b^2-a^2)x^2=a^2b^2-a^2c^2=a^2(b^2-c^2).(b2−a2)x2=a2b2−a2c2=a2(b2−c2).

Since

b2−c2=b2−(a2−b2)=2b2−a2,b^2-c^2=b^2-(a^2-b^2)=2b^2-a^2,b2−c2=b2−(a2−b2)=2b2−a2,

this is messy. Instead use c2=a2−b2⇒a2−c2=b2c^2=a^2-b^2 \Rightarrow a^2-c^2=b^2c2=a2−b2⇒a2−c2=b2, so

a2b2−a2c2=a2(b2−c2).a^2b^2-a^2c^2=a^2(b^2-c^2).a2b2−a2c2=a2(b2−c2).

A cleaner elimination is:

From ellipse,

x2+a2b2y2=a2.x^2+\frac{a^2}{b^2}y^2=a^2.x2+b2a2​y2=a2.

Subtract circle equation x2+y2=c2x^2+y^2=c^2x2+y2=c2:

(a2b2−1)y2=a2−c2.\left(\frac{a^2}{b^2}-1\right)y^2=a^2-c^2.(b2a2​−1)y2=a2−c2.

Now,

a2b2−1=a2−b2b2=c2b2,\frac{a^2}{b^2}-1=\frac{a^2-b^2}{b^2}=\frac{c^2}{b^2},b2a2​−1=b2a2−b2​=b2c2​,

and

a2−c2=b2.a^2-c^2=b^2.a2−c2=b2.

Therefore,

c2b2y2=b2⇒c2y2=b4⇒y2=b4c2.\frac{c^2}{b^2}y^2=b^2 \quad\Rightarrow\quad c^2y^2=b^4 \quad\Rightarrow\quad y^2=\frac{b^4}{c^2}.b2c2​y2=b2⇒c2y2=b4⇒y2=c2b4​.
  1. Use the given area of triangle PF1F2PF_1F_2PF1​F2​

The base F1F2F_1F_2F1​F2​ has length

F1F2=2c.F_1F_2=2c.F1​F2​=2c.

Since the foci lie on the xxx-axis, the perpendicular distance of P(x,y)P(x,y)P(x,y) from the line F1F2F_1F_2F1​F2​ is ∣y∣|y|∣y∣. So the area of triangle PF1F2PF_1F_2PF1​F2​ is

12⋅2c⋅∣y∣=c∣y∣.\frac12\cdot 2c\cdot |y| = c|y|.21​⋅2c⋅∣y∣=c∣y∣.

Given area =30=30=30, we get

c∣y∣=30.c|y|=30.c∣y∣=30.

Using

y2=b4c2⇒∣y∣=b2c,y^2=\frac{b^4}{c^2} \quad\Rightarrow\quad |y|=\frac{b^2}{c},y2=c2b4​⇒∣y∣=cb2​,

we have

c⋅b2c=b2=30.c\cdot \frac{b^2}{c}=b^2=30.c⋅cb2​=b2=30.

Thus,

b2=30.b^2=30.b2=30.
  1. Use the major axis length

The length of the major axis is 2a=172a=172a=17, so

a=172.a=\frac{17}{2}.a=217​.

Then

a2=2894.a^2=\frac{289}{4}.a2=4289​.

Now,

c2=a2−b2=2894−30=289−1204=1694.c^2=a^2-b^2=\frac{289}{4}-30=\frac{289-120}{4}=\frac{169}{4}.c2=a2−b2=4289​−30=4289−120​=4169​.

Hence,

c=132.c=\frac{13}{2}.c=213​.

Therefore the distance between the foci is

2c=13.2c=13.2c=13.
  1. Check options

The correct option is

13\boxed{13}13​

which is Option C.

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