If the points of intersection of two distinct conics and lie on the curve , then times the area of the rectangle formed by the intersection points is _________. [2024]

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Solution

Q.
If the points of intersection of two distinct conics $x^{2} + y^{2} = 4 b$ and $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ lie on the curve $y^{2} = 3 x^{2}$ , then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is _________. [2024]

Ans.
(432)

We have, $x^{2} + y^{2} = 4 b$ ... (i)

and $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$ ... (ii)

From (i) and (ii), $x^{2} = \frac{16 b}{b + 4}$ and $y^{2} = \frac{4 b^{2}}{b + 4}$

$∵$ These points lie on curve $y^{2} = 3 x^{2}$ .

So, $\frac{4 b^{2}}{b + 4} = 3 \times \frac{16 b}{b + 4}$

$\Rightarrow b^{2} = 12 b \Rightarrow b (b - 12) = 0$

$\Rightarrow b = 12 (b \neq 0)$

So, the points are $(\pm 2 \sqrt{3}, \pm 6)$

$∴$ Area of rectangle = $4 \sqrt{3} \times 12 = 48 \sqrt{3}$ sq. units

Thus, $3 \sqrt{3}$ times of the area of rectangle $= 3 \sqrt{3} \times 48 \sqrt{3} = 432$ .

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