Find the equation of normal at a point on the curve x² = 4y, which passes through the point (1, 2). Also, find the equation of the corresponding tangent.

Let the tangent and normal passes through point (a, b) as shown

The red line is normal which is passing through (1, 2) and (a, b) and the blue line is tangent to curve x² = 4y at (a, b)

The slope of tangent at (a, b) is given by let it be denoted as m_t

let us find slope of tangent at (a, b)

⇒ 4y = x²

Differentiate with respect to x

We have to find at (a, b)

Now let the slope of normal be m_n

As tangent and normal are perpendicular to each other their product of slopes is -1

⇒ m_tm_n = -1

Substitute

As the slope of normal is and point on normal is (1, 2) hence equation of normal in slope point form

As (a, b) lies on normal

⇒ ab = 2

As (a, b) also lies on the curve x² = 4y

⇒ a² = 4b

Multiply by b² on both sides

⇒ (ab)² = 4b³

Since ab = 2

⇒ 4 = 4b³

⇒ b = 1

Put b = 1 in ab =2 we get a = 2

Hence (a, b) = (2, 1)

Put these values of (a, b) in (p) to get equation of normal

⇒ y – 2 = - x + 1

⇒ x + y = 3

Put (a, b) value in to get slope of tangent

⇒ m_t = 1

Slope of tangent is 1 and is passing through (a, b) that is (2, 1), hence writing equation in slope point form

⇒ y – 1 = 1(x – 2)

⇒ x – y = 1

Hence equation of tangent is x – y = 1 and normal is x + y = 3