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
=0 Now by differentiating equation of curve
x2+y2−2x−4y+1=0 with respect to y we get,
⇒2x.+2y−2.−4=0 ⇒= ⇒[](x1,y1)= ∵[](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)