# A tangent having slope –12 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 - Mathematics and Statistics

Sum

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

Is there an error in this question or solution?