Area of a triangle formed by tangent and normal to the curve x² a² + y² b² =1 at p ( a 2 , b 2 ) with the x -axis is

Step-by-step solution

Given curve x² a² + y² b² =1 and point P ( a 2 , b 2 ) Slope of tangent at P is -b / a . Slope of normal at P is a / b . Equation of tangent on the curve b x+a y= 2 a b Equation of normal on the tangent of the curve -a x+b y= (b²-a² ) 2 Equation of x -axis y=0 Solving these equations in successive manner, we get three points of triangle ( a²-b² a 2 , 0 ),( 2 a, 0) and ( a 2 , b 2 ) Now, area of triangle = 1 2 | array ccc a²-b² a 2 & 0 & 1 a / 2 a & 0 / 2 & 1 1 array |= -b 2 2 a²-b² a 2 - 2 a = -b 4 a (-a²-b² )= b (