First, find the derivative of the function.
#y = sqrt(3 + x^2)#
Letting #y = sqrt(u)# and #u = 3 + x^2#.
#y' = 1/(2u^(1/2))#
#u' = 2x#
#dy/dx = 1/(2u^(1/2)) xx 2x#
#dy/dx = (2x)/(2(3 + x^2)^(1/2))#
#dy/dx = x/(3 + x^2)^(1/2)#
Now, we must find the slope of the line #x - 2y = 1#, since if the tangent is parallel to this line, then the tangent will have an equal slope.
#x - 2y = 1 -> -2y = 1 - x -> y = -1/2 + 1/2x#
In slope intercept form, #y = mx + b#, the slope is given by #m#. Hence, the slope is #1/2#.
Because the slope of the tangent is given by evaluating #x = a# inside the derivative, we can use the slope as y and solve for our point, #x#.
#1/2 = x/sqrt(3 + x^2)#
#1/2sqrt(3 + x^2) = x#
#1/4(3 + x^2) = x^2#
#3/4 + 1/4x^2 = x^2#
#3/4 = x^2 - 1/4x^2#
#3/4 = 3/4x^2#
#(3/4)/(3/4) = x^2#
#1 = x^2#
#x = +-1#
So, there are two possible equations of tangents. Let's next find the y coordinate of each point. This can be obtained by inserting #x = a# into the original function.
#y = sqrt(3 + 1^2)" AND "y = sqrt(3 + (-1)^2)#
#y = 2" AND "y = 2#
We can now use the slope and the points to find the equation of each tangent.
tangent 1: Where #m = 1/2# and the line passes through #(1, 2)#
#y - y_1 = m(x - x_1)#
#y - 2 = 1/2(x - 1)#
#y - 2 = 1/2x - 1/2#
#y = 1/2x + 3/2#
Tangent 2: Where #m = 1/2# and the line passes through #(-1, 2)#
#y - y_1 = m(x - x_1)#
#y - 2 = 1/2(x + 1)#
#y- 2= 1/2x + 1/2#
#y = 1/2x +5/2#
Hopefully this helps!
