If the points of intersection of two distinct conics x²+y²=4b and x²/16+y²/b²=1 lie on the curve y²=3x², then 3√3 times the area of the rectangle formed by the intersection points is _________.

Solution

Maths sir · 5 steps

Put y² = 3x² into both conics, solve for b, and throw out the value that makes them the same curve.

1. Use the circle. On y² = 3x², the circle gives x² + 3x² = 4b, so x² = b and y² = 3b at every intersection point.
2. Use the ellipse. Put x² = b and y² = 3b into the ellipse: b/16 + 3b/b² = 1, which is b/16 + 3/b = 1.

   b² - 16b + 48 = 0 ⇒ (b-4)(b-12) = 0

3. Reject b = 4. With b = 4 the ellipse becomes x² + y² = 16, which is exactly the circle, but the conics must be distinct. So b = 12.
4. Find the rectangle. Now x² = 12 and y² = 36, so the points are (± 2√3, ± 6). The sides are 4√3 and 12, so the area is 48√3.
5. So. 3√3 × 48√3 = 48 × 9 = 432.

The common slip: 144. Keeping b = 4 makes the two conics the same circle; it gives points (± 2, ± 2√3) and an answer of 144.

The idea to keep. When intersection points lie on a known curve, substitute that curve into both equations to get a condition on the parameter. Then check every root against the conditions given, like "distinct".

Answer from the NTA final answer key. Solution by PrepWiser.