How do you find the equation of tangent line to the curve #y= sqrt(3+x^2)# that is parallel to the line x-2y=1?

Question

Charles Arocha · Accepted Answer

There are two possible equations:
•#y = 1/2x+ 3/2#

•#y = 1/2x + 5/2# 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!

Emily Ammerman · Answer

undefined To find the equation of the tangent line to the curve y = sqrt(3 + x^2) that is parallel to the line x - 2y = 1, we need to follow these steps:

1. Differentiate the equation of the curve y = sqrt(3 + x^2) with respect to x to find the derivative dy/dx.
2. Set the derivative dy/dx equal to the slope of the given line x - 2y = 1, which is 1/2.
3. Solve the resulting equation for x to find the x-coordinate(s) of the point(s) where the tangent line is parallel to the given line.
4. Substitute the x-coordinate(s) obtained in step 3 into the original equation y = sqrt(3 + x^2) to find the corresponding y-coordinate(s).
5. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the tangent line and m is the slope, to write the equation of the tangent line using the coordinates found in step 4.

By following these steps, you will obtain the equation of the tangent line to the curve y = sqrt(3 + x^2) that is parallel to the line x - 2y = 1.