Sum

A tangent having slope `–1/2` to the ellipse 3x^{2} + 4y^{2} = 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

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#### Solution

The equation of the ellipse is 3x^{2} + 4y^{2} = 12

i.e.`x^2/4 + y^2/3` = 1

Comparing with `x^2/"a"^2 + y^2/"b"^2` = 1, we get

a^{2} = 4, b^{2} = 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)` ...[∵ a^{2} = 4, b^{2} = 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

Concept: Conic Sections - Ellipse

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