Q84.Let S be the focus of the hyperbola x23 −y25 = 1 A(√6, √5) and passing through the point S . If O is the origin and SAB is a diameter of C , then the square of the area of the triangle OSB is equal to___________
What This Question Tests
This question tests the ability to find the focus of a hyperbola, use the midpoint formula to find coordinates of a point given a diameter, and then calculate the area of a triangle formed by these points and the origin. The question has slight formatting issues but is interpretable based on common JEE problem structures.
Concepts Tested
Formulas Used
e^2 = 1 + b^2/a^2
Foci (±ae, 0)
Midpoint: ((x1+x2)/2, (y1+y2)/2)
Area of triangle: 1/2 |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)| or 1/2 |x_S y_B - x_B y_S| for O(0,0)
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📋 Question Details
- Chapter
- Hyperbola
- Topic
- Equation of hyperbola, foci, properties of diameter, area of triangle
- Year
- 2024
- Shift
- 08 Apr Shift 2
- Q Number
- Q84
- Type
- Numerical
- NCERT Ref
- Class 11 Mathematics Ch 11: Conic Sections
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