Given curve is y \(=\frac{x^2+3}{x-1}\)
Slope of tangent to given curve is m = dy/dx \(=\frac{(x-1)\times2x-(x^2+3)\times1}{(x-1)^2}\)
Slope of tangent to curve at x = 2 is m \(=\frac{1\times4-7}{1^2}=\frac{4-7}{1}\) = -3
Ordinate of point P on curve when x = 2 is y \(=\frac{2^2+3}{2-1}\) = 7/1 = 7
Hence, point (2, 7) lies on curve.
Thus tangent of given curve has slope -3 and passing through point (2, 7) when x = 2.
∴ Equation of tangent at x = 2 is y - y1 = m(x - x1)
⇒ y - 7 = -3(x - 2) (∵ y1 = 7, x1 = 2, m = -3)
⇒ y - 7 = -3x+6
⇒ 3x+y-7-6 = 0
⇒ 3x+y = 13
Hence, equation of required tangent is 3x+y = 13.