UPSC NDA Math Model Paper 3

Given: Equation of curve is x2+y2−2x−4y+1=0 and tangent to the given curve is parallel to Y - axis.
Let point of the contact be (x1,y1)
As we know that, if a tangent of any curve is parallel to Y− axis then

dx

dy

=0
Now by differentiating equation of curve x2+y2−2x−4y+1=0 with respect to y we get,
⇒2x.

dx

dy

+2y−2.

dx

dy

−4=0
⇒

dx

dy

=

2−y

x−1

⇒[

dx

dy

](x1,y1)=

2−y1

x1−1

∵[

dx

dy

](x1,y1)=0
⇒y1=2
∵(x1,2) is the point of contact i.e x=x1,y=2 will satisfy the equation x2+y2−2x−4y+1=0
i.e x12+4−2x1−8+1=0
⇒x12−2x1−3=0
⇒(x1−3).(x1+1)=0
⇒x1=−1 or 3
So, the required points are: (- 1, 2) and (3, 2)