A tangent having slope - 1 2 to the ellipse 3x^2 + 4y^2 = 12 intersects the X-axis and Y-axis at the points A and B respectively. if O is the origin, then the area of AOB is ...

Step-by-step solution

The equation of the given ellipse can be rewritten as x^2 4 + y^2 3 = 1 Here, a^2 = 4 and b^2 = 3 The equation of a tangent to the ellipse with slope m is given by y = mx a^2m^2 + b^2 Given the slope m = - 1 2 , substituting the values we get: y = - 1 2 x 4 (- 1 2 )^2 + 3 y = - 1 2 x 1 + 3 y = - 1 2 x 2 x + 2y = 4 The tangent intersects the X-axis at A( 4, 0) and the Y-axis at B(0, 2) The area of AOB is 1 2 OA OB = 1 2 4 2 = 4 sq. units Answer: 4 sq. units