How do you find the equation of the tangent and normal line to the curve
#y=1/x# at x=2?

Question

How do you find the equation of the tangent and normal line to the curve 
#y=1/x# at x=2?

Christopher Agresta · Accepted Answer

Step 1 - Find the value of #y# when #x=2#:
#x=2=>y=1/2# Step 2 - Find the gradient of the tangent at #x=2#: #y=1/x#, #:. y=x^-1#
So, #(dy)/(dx)=(-1)x^-2=-1/x^2# When #x=2=> (dy)/(dx)=-1/2^2=-1/4# Step 3 - Eq'n of the Tangent when #x=2#:
The tangent passes through #(2,1/2)# and has gradient #-1/4#
Using #y-y_1=m(x-x_1)# we have the equation of the tangent is given by:
#y-1/2=-1/4(x-2)#
#:. 4y-2=-1(x-2)#  (multiplying by 4)
#:. 4y-2=2-x#
#:. 4y+x=4# Step 4 - Eq'n of the Normal when #x=2#:
Normal is perpendicular to Tangent, so
gradient of Normal = #-1/(-1/4)=4# The normal passes through #(2,1/2)# and has gradient #4#
Using #y-y_1=m(x-x_1)# we have the equation of the normal is given by:
#y-1/2=4(x-2)#
#:. 2y-1=8(x-2)#  (multiplying by 2)
#:. 2y-1=8x-16#
#:. 2y-8x=-15#

Emma Aldridge · Answer

undefined To find the equation of the tangent line to the curve $y = \frac{1}{x}$ at $x = 2$, we need to find the derivative of the function $y$ with respect to $x$, then evaluate it at $x = 2$ to find the slope of the tangent line. Once we have the slope, we can use the point-slope form of a line to find the equation of the tangent line.

First, let's find the derivative of $y$ with respect to $x$:
$$y = \frac{1}{x}$$
$$y' = -\frac{1}{x^2}$$

Now, evaluate the derivative at $x = 2$:
$$y'(2) = -\frac{1}{2^2} = -\frac{1}{4}$$

So, the slope of the tangent line at $x = 2$ is $-\frac{1}{4}$.

Now, we need to find the $y$-coordinate corresponding to $x = 2$ by plugging $x = 2$ into the original function:
$$y(2) = \frac{1}{2} = 0.5$$

Now that we have the slope and a point on the tangent line, we can use the point-slope form of a line to find the equation of the tangent line:
$$y - y_1 = m(x - x_1)$$
$$y - 0.5 = -\frac{1}{4}(x - 2)$$

This is the equation of the tangent line.

To find the equation of the normal line, we use the fact that the normal line is perpendicular to the tangent line. Therefore, the slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is the negative reciprocal of $-\frac{1}{4}$, which is $4$.

Using the point-slope form with the same point $(2, 0.5)$, we find the equation of the normal line:
$$y - 0.5 = 4(x - 2)$$

This is the equation of the normal line.

How do you find the equation of the tangent and normal line to the curve #y=1/x# at x=2?