Find the points on the curve `x^2+y^2-2x-3=0`at which the tangents are parallel to the x-axis.

Options

A. (1,2) and (1,-2)
B. `(0,sqrt(3)) and (0,-sqrt(3))`
C. (3,0) and (-3,0)
D. (2,1) and (2,-1)

Answer

A

Explanation

Given equation curve is
`x^(2)+y^(2)-2x-3=0`
Oln differntiatin we get
`2x+2y (dy)/(dx)-2=0`
`rarr y (dy)/(dx)=-x+1`
`rarrr (dy)/(dx)=-(x+1)/(y)`
since tangent is parallel to x axis
`therefore (dy)/(dx)=0 rarr -(x+1)/(y) = 0 rarr x=1 `
`therefore ` from equation (1) we have
`1+y^(2)-2-3 =0 rarr y =pm 2`
Hence required points ae (1,2) and (1,-2)`