The centre of a circle C is at the centre of the ellipse E : $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$. Let C pass through the foci $$F_1$$ and $$F_2$$ of E such that the circle C and the ellipse E intersect at four points. Let P be one of these four points. If the area of the triangle $$PF_1F_2$$ is 30 and the length of the major axis of E is 17, then the distance between the foci of E is :

Let the centre of the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ be the origin $$O(0,0)$$.
Because the length of the major axis is given as 17,

$$2a = 17 \; \Longrightarrow \; a = \dfrac{17}{2}$$

For an ellipse, the distance of each focus from the centre is $$c$$ where

$$c^{2}=a^{2}-b^{2} \quad -(1)$$

The foci are therefore $$F_{1}(-c,0)$$ and $$F_{2}(c,0)$$.

The circle $$C$$ is also centred at $$O$$ and passes through the foci, hence its radius is $$c$$ and its equation is

$$x^{2}+y^{2}=c^{2} \quad -(2)$$

Let $$P(x,y)$$ be one of the four common points of the ellipse and the circle.
Then $$P$$ satisfies both $$(2)$$ and the ellipse equation, giving

$$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1 \quad -(3)$$

From $$(2)$$: $$y^{2}=c^{2}-x^{2} \quad -(4)$$

Substitute $$(4)$$ into $$(3)$$:

$$\dfrac{x^{2}}{a^{2}}+\dfrac{c^{2}-x^{2}}{b^{2}}=1$$

Simplify the left side:

$$x^{2}\left(\dfrac{1}{a^{2}}-\dfrac{1}{b^{2}}\right)+\dfrac{c^{2}}{b^{2}}=1$$

Using $$(1)$$, $$b^{2}=a^{2}-c^{2}$$, so

$$\dfrac{1}{a^{2}}-\dfrac{1}{b^{2}}=\dfrac{b^{2}-a^{2}}{a^{2}b^{2}}=-\dfrac{c^{2}}{a^{2}b^{2}}$$

Hence

$$-\dfrac{c^{2}}{a^{2}b^{2}}\,x^{2}+\dfrac{c^{2}}{b^{2}}=1$$

Multiply by $$b^{2}$$:

$$c^{2}-\dfrac{c^{2}}{a^{2}}\,x^{2}=b^{2}$$

Replace $$b^{2}$$ again by $$a^{2}-c^{2}$$:

$$c^{2}-\dfrac{c^{2}}{a^{2}}\,x^{2}=a^{2}-c^{2}$$

Rearrange to solve for $$x^{2}$$:

$$\dfrac{c^{2}}{a^{2}}\,x^{2}=2c^{2}-a^{2}$$

$$x^{2}=a^{2}\dfrac{2c^{2}-a^{2}}{c^{2}} \quad -(5)$$

Now obtain $$y^{2}$$ from $$(4)$$:

$$y^{2}=c^{2}-x^{2}=c^{2}-a^{2}\dfrac{2c^{2}-a^{2}}{c^{2}} \quad -(6)$$

Next use the area condition.
For the triangle with vertices $$P, F_{1}(-c,0), F_{2}(c,0)$$:

• Base $$F_{1}F_{2}=2c$$.
• The altitude from $$P$$ to the base is $$|y|$$ (since the base lies on the $$x$$-axis).

Hence

$$\text{Area}=\dfrac{1}{2}\times 2c \times |y|=c|y|$$

The area is given as 30, so

$$c|y| = 30 \;\Longrightarrow\; y^{2}=\dfrac{900}{c^{2}} \quad -(7)$$

Equate $$(6)$$ and $$(7)$$:

$$\dfrac{900}{c^{2}} = c^{2}-a^{2}\dfrac{2c^{2}-a^{2}}{c^{2}}$$

Multiply by $$c^{2}$$ to clear denominators:

$$900 = c^{4}-a^{2}(2c^{2}-a^{2}) \quad -(8)$$

Insert $$a^{2}=\left(\dfrac{17}{2}\right)^{2}=\dfrac{289}{4}$$ into $$(8)$$:

$$900 = c^{4}-\dfrac{289}{4}\left(2c^{2}-\dfrac{289}{4}\right)$$

Multiply by 16 to avoid fractions:

$$16c^{4}-8\!\cdot\!289\,c^{2}+289^{2}-14400=0$$

That is

$$16c^{4}-2312c^{2}+69121=0$$

Let $$k=c^{2}$$. The quadratic in $$k$$ becomes

$$16k^{2}-2312k+69121=0$$

Using the quadratic formula,

$$k=\dfrac{2312\pm\sqrt{2312^{2}-4\!\cdot\!16\!\cdot\!69121}}{32}$$

The discriminant is

$$2312^{2}-64\!\cdot\!69121 = 5345344-4423744 = 921600 = 960^{2}$$

Hence

$$k=\dfrac{2312\pm960}{32}$$

This gives two values: $$k=102.25$$ and $$k=42.25$$.
But $$k=c^{2}$$ must satisfy $$c^{2}\lt a^{2}=72.25$$, so we take

$$c^{2}=42.25 \;=\;\left(6.5\right)^{2}$$

Therefore

$$c = 6.5, \quad 2c = 13$$

The distance between the foci of the ellipse is 13.
Hence the correct option is Option B (13).