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 ______

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Ellipse

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

33

times the area of the rectangle formed by the intersection points is ______

Answer: 432

Substituting

y^{2} = 3 x^{2}

into

x^{2} + y^{2} = 4 b

4 x^{2} = 4 b ⟹ x^{2} = b

y^{2} = 3 b

.

Substituting into

\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1

\frac{b}{16} + \frac{3}{b} = 1 ⟹ b^{2} - 16 b + 48 = 0 ⟹ b = 4 or 12

b = 4

makes the two conics coincide (not distinct), so

b = 12

.

Then

x^{2} = 12

y^{2} = 36

, giving intersection points

(\pm 23, \pm 6)

-- the vertices of a rectangle of width

43

and height

12

Area = 43 \times 12 = 483

33 \times Area = 33 \times 483 = 3 \times 48 \times 3 = 432

Key confirmed: 432.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer