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Question
A tangent having slope `–1/2` to the ellipse 3x2 + 4y2 = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle
Solution
The equation of the ellipse is 3x2 + 4y2 = 12
i.e.`x^2/4 + y^2/3` = 1
Comparing with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = 4, b2 = 3
The equation of tangent with slope m is
y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)`
i.e., y = `"m"x ± sqrt(4"m"^2 + 3)` ...[∵ a2 = 4, b2 = 3]
∴ y = `-1/2x ± sqrt(4(1/4) + 3) ...[because "m" = -1/2]`
∴ y = `-x/2 ± 2`
∴ x + 2y ± 4 = 0 ...(1)
It meets X axis at A
∴ for A, put y = 0 in equation (1), we get,
x = ±4
∴ A = (±4, 0)
Similarly, B = (0, ±2)
∴ OA = 4, OB = 2
∴ area of ΔOAB = `1/2*"OA"*"OB"`
= `1/2*4*2`
= 4 sq. units
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