UPSC NDA Math Model Paper 5

Given curve x2=4y
Differentiating w.r.t x on both sides
⇒

d

dx

x2=

d

dx

4y
⇒2x=4

dy

dx

Let the point on which normal cuts the curve be
(x1,y1)
⇒

dy

dx

=

x1

2

⇒ Slope of normal m=−

dx

dy

=−

2

x1

⇒

2−y1

1−x1

=−

2

x1

Also y1=

x2

4

⇒(2−

x12

4

)x1=2x1−2
⇒2x1−

x13

4

=2x1−2
⇒x13=8⇒x1=2
y1=

x12

4

=1
Now equation of the normal passing through (2, 1) and (1, 2)

y−y1

x−x1

=

y2−y1

x2−x1

⇒

y−1

x−2

=

2−1

1−2

⇒

y−1

x−2

=−1
⇒y−1=2−x
⇒y+x−3=0
So, distance of the origin (0, 0) from normal is;
D=|

ap+bq+c

√a2+b2

|
⇒D=|

1×0+1×0+(−3)

√12+12

|
⇒D=|

−3

√2

|=

3

√2