Ellipse
Intersection of Conics — Area of Rectangle
nta_pyq_2024_jan
Grade 11
Question:
If the points of intersection of two distinct conics $x^2+y^2=4b$ and $\dfrac{x^2}{16}+\dfrac{y^2}{b^2}=1$ lie on the curve $y^2=3x^2$, then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is
Step-by-Step Solution
Key Concept: Substitute $y^2=3x^2$ into both conics to find $x^2$ (and hence $b$). Then determine the four intersection points and compute the area of the rectangle they form. Multiply by $3\sqrt{3}$.
$b=12$. Points $(\pm\sqrt{12},\pm6)$ form rectangle with area $2\sqrt{12}\times12=48\sqrt{3}$. $3\sqrt{3}\times48\sqrt{3}=432$.
Correct Answer: 432