UPSC NDA Math Model Paper 6

Given curve y2=16x
Differentiating w.r.t x on both sides
⇒

d

dx

y2=

d

dx

16x
⇒2y

dy

dx

=16
The point on which tangent touches the curve is (1,4)
⇒

dy

dx

=

8

4

=2
⇒ Slope of tangent m=2
Now equation of the tangent:
(y−y1)=m(x−x1)
⇒y−4=2(x−1)
⇒y−2x−2=0
So, distance of the origin (0, 0) from normal is;
D=|

ap+bq+c

√a2+b2

|
⇒D=|

1×0+(−2)×0+(−2)

√12+(−2)2

|
⇒D=|

−2

√5

|=

2

√5