Question:
Let the line \( y = x + 3 \) intersect the parabola \( x^2 = 4y \) at the points \( A \) and \( B \). If \( F \) is the focus of the parabola, then the area of triangle \( AFB \) is :
Let the line \( y = x + 3 \) intersect the parabola \( x^2 = 4y \) at the points \( A \) and \( B \). If \( F \) is the focus of the parabola, then the area of triangle \( AFB \) is :
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Find A and B by solving the line with the parabola, take the focus F(0, 1), then use the triangle area formula.
Updated On: Nov 12, 2026
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The Correct Option is C
Solution and Explanation
Substitute \( y = x + 3 \) in \( x^2 = 4y \): \( x^2 - 4x - 12 = 0 \), so \( (x - 6)(x + 2) = 0 \) and \( x = 6, -2 \).
Hence \( A(6, 9) \) and \( B(-2, 1) \).
For \( x^2 = 4y \), \( a = 1 \), so the focus is \( F(0, 1) \).
Area = \( \frac{1}{2} |6(1 - 1) + (-2)(1 - 9) + 0(9 - 1)| = \frac{1}{2} \times 16 = 8 \).
Answer: (3) 8
Hence \( A(6, 9) \) and \( B(-2, 1) \).
For \( x^2 = 4y \), \( a = 1 \), so the focus is \( F(0, 1) \).
Area = \( \frac{1}{2} |6(1 - 1) + (-2)(1 - 9) + 0(9 - 1)| = \frac{1}{2} \times 16 = 8 \).
Answer: (3) 8
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