Show that at any point ( x , y ) on the curve y = b e x a , the length of the sub- tangent is constant and the length of the sub-normal is y 2 a .

Let P ( x , y ) be a point on the curve. Equation of the curve is y = b e x a ⇒ dy dx = b . e x a · 1 a = y a = m Length of the sub-tangent y 1 m = y y a = a = constant Length of the sub-normal y 1 m = y · y a = y 2 a ,

Find the angle between the curves x + y + 2 = 0 ; x 2 + y 2 - 10 y = 0 \text { 1ind the angle between the curves givan below: }\text { 1ind the angle between the curves givan below: }1ind the angle between the curves givan below:1ind the angle between the curves givan below:lini the angle between the curves given belowelini the angle between the curves given beloweFind the angle between the curves given below:

Given equation of line is x + y + 2 = 0 ⇒ x = - ( y + 2 ) ......(i) Equation of the curve x 2 + y 2 - 10 y = 0 ......(ii) Solving equation (i) and (ii), we get ( y + 2 ) 2 + y 2 - 10 y = 0 ⇒ y 2 + 4 y + 4 + y 2 - 10 y = 0 ⇒ 2 y 2 - 6 y + 4 = 0 ⇒ y 2 - 3 y + 2 = 0 ⇒ ( y + 1 ) ( y - 2 ) = 0 ⇒ y = 1 or y = 2 When y = 1 , ⇒ x = - 1 + 2 = - 3 (from equation(i)) When y = 2 , ⇒ x = - ( 2 + 2 ) = - 4 (from equation(i)) The points of intersection are P ( - 3 , 1 ) and Q ( - 4 , 2 ) Now, differentiate equation(ii) w.r.to x . ⇒ 2 x + 2 y d y d x - 10 d y d x = 0 ⇒ 2 dy dx ( y - 5 ) = - 2 x ⇒ dy dx = - x y - 5 From equation (i), compare with the y = m x + c Slope is m 2 = - 1 Case(i): Slope m 1 = d y d x at P = - - 3 1 - 5 = - 3 4 and slope m 2 = - 1 . Let θ be the angle between the curves, then tanθ = m 1 - m 2 1 + m 1 m 2 = - 3 4 + 1 1 + 3 4 = 1 7 Case(ii): Slope m 1 = d y d x at Q - 4 2 - 5 = - 4 3 and slope m 2 = - 1 . Let θ be the angle between the curves, then tanθ = m 1 - m 2 1 + m 1 m 2 = - 4 3 + 1 1 + 4 3 = 1 7 ∴ θ = tan - 1 1 7