A function is defined parametrically by the equations x = 2 t + t 2 sin 1 t , t ≠ 0 0 , t = 0 and y = 1 t sin t 2 , t ≠ 0 0 , t = 0 Find the equation of the tangent and normal at the point for t = 0 if they exist.

At t = 0 , x = 0 , y = 0 i.e. the point is origin 0 , 0 . Using first principle of differentiation, d x d t = lim t → 0 f ( t ) - f ( 0 ) t - 0 = lim t → 0 2 t + t 2 sin 1 t t = 2 + lim t → 0 t sin 1 t = 2 (since sin 1 t is a fixed but oscillating number at t = 0 . Similarly, d y d t = lim t → 0 g ( t ) - g ( 0 ) t - 0 = lim t → 0 1 t sin t 2 t = lim t → 0 sin t 2 t 2 = lim t 2 → 0 sin t 2 t 2 = 1 Now, d y d x t = 0 = d y d t t = 0 d x d t t = 0 = 1 2 So, slope of tangent = d y d x t = 0 = 1 2 and slope of normal = - 1 d y d x t = 0 = - 2 Thus equation of tangent is y - 0 = 1 2 ( x - 0 ) i.e. x = 2 y and equation of normal is y - 0 = - 2 ( x - 0 ) i.e. 2 x + y = 0

Find the angle of intersection between the curves y 2 = 2 x π and y = sin x at x = π 2 .

Angle of intersection between curves is determined by the angle between tangents to them at the point of intersection. Given x = π 2 , putting it in given curves y 2 = 2 x π & y = sin x , we get y = 1 , so point of intersection is π 2 , 1 . Now, y 2 = 2 x π ⇒ 2 y y ' = 2 π ⇒ y ' π 2 , 1 = 1 π . . . ( 1 ) y = sin x ⇒ y ' = cos x ⇒ y ' π 2 , 1 = 0 . . . ( 2 ) Let angle of intersection i.e. angle between tangents at point of intersection be θ , then tan θ = 1 π - 0 1 + 1 π · 0 = 1 π ⇒ θ = tan - 1 1 π ⇒ θ = cot - 1 π

Show that the angle between the tangent at any point ' A ' of the curve ln x 2 + y 2 = C tan - 1 y x and the line joining A to the origin is independent of the position of A on the curve.

Let the point is A x 1 , y 1 . Slope of line joining ( 0 , 0 ) & x 1 , y 1 is m 1 = y 1 x 1 . We have ln x 2 + y 2 = C tan - 1 y x . Differentiating both sides, ⇒ ( 2 x + 2 y y ' ) x 2 + y 2 = C x y ' - y x 2 1 + y 2 x 2 ⇒ ( 2 x + 2 y y ' ) = C x y ' - y ⇒ y ' = 2 x + C y C x - 2 y ⇒ y ' x 1 , y 1 = 2 x 1 + C y 1 C x 1 - 2 y 1 i.e. slope of tangent at A , m 2 = 2 x 1 + C y 1 C x 1 - 2 y 1 . Now, required acute angle = tan - 1 m 1 - m 2 1 + m 1 m 2 = tan - 1 y 1 x 1 - 2 x 1 + C y 1 C x 1 - 2 y 1 1 + y 1 x 1 · 2 x 1 + C y 1 C x 1 - 2 y 1 = tan - 1 C x 1 y 1 - 2 y 1 2 - 2 x 1 2 - C x 1 y 1 C x 1 2 - 2 x 1 y 1 + 2 x 1 y 1 + C y 1 2 = tan - 1 - 2 C , which is independent of the coordinates of A . Thus the required angle is independent of the position of A .

Find the condition that the curves x 2 a + y 2 b = 1 & x 2 a ' + y 2 b ' = 1 may cut orthogonally.

Curves cut orthogonally if the tangents at their point of intersection are perpendicular. Finding point of intersection: Given curves x 2 a + y 2 b = 1 & x 2 a ' + y 2 b ' = 1 ⇒ x 2 = a 1 - y 2 b . Putting this value in other curve, we get a a ' 1 - y 2 b + y 2 b ' = 1 ⇒ y 2 = b b ' a - a ' a b ' - a ' b . So, x 2 = a a ' b ' - b a b ' - a ' b So, point of intersection is a a ' b ' - b a b ' - a ' b , b b ' a - a ' a b ' - a ' b Now, x 2 a + y 2 b = 1 ⇒ 2 x a + 2 y y ' b = 0 ⇒ y ' = - b x a y So, slope of tangent at the point of intersection will be m 1 = - b a · a a ' b ' - b b b ' a - a ' = - b a ' b ' - b a b ' a - a ' . . . ( 1 ) Similarly, x 2 a ' + y 2 b ' = 1 ⇒ 2 x a ' + 2 y y ' b ' = 0 ⇒ y ' = - b ' x a ' y . So, slope of tangent at the point of intersection will be m 2 = - b ' a ' · a a ' b ' - b b b ' a - a ' = - a b ' b ' - b b a ' a - a ' . . . ( 2 ) For orthogonal intersection, these tangents have to be perpendicular, thus m 1 m 2 = - 1 ⇒ - b a ' b ' - b a b ' a - a ' · - a b ' b ' - b b a ' a - a ' = - 1 ⇒ b ' - b a - a ' = - 1 ⇒ a - b = a ' - b '