Question

The normal to the curve x² = 4y passing (1,2) is
A. x + y = 3
B. x – y = 3
C. x + y = 1
D. x – y = 1

Answer 1

Answer:

ANSWER

Given, x

2

=4y

Slope of tangent to the curve will be 4

dx

dy

=2x

⇒

dx

dy

=

2

x

So slope of normal will be −

dy

dx

=

x

−2

Let the point on the curve be (h,k)

So slope of normal will be

h

−2

Equation of normal is given by (y−k)=

h

−2

(x−h)

It is given that the normal passes through (1,2)

⇒(2−k)=

h

−2

(1−h).............(1)

Also, (h,k) lies on the curve so h

2

=4k

From (1) we have

4

h

3

=2h+2−2h=2

⇒h

3

=8

⇒h=2

k=

4

h

2

⇒k=1

So the equation will be

y−1=

2

−2

(x−2)

x+y=3