Tangent and Normal Lines
Solved Problems
Example 9.
Find the equation of the normal to the graph of the function \[y = \frac{{x + 1}}{{x - 1}}\] at the point where \(x = 2.\)
Solution.
Differentiate the given function using the quotient rule:
\[ y^\prime = \left( {\frac{{x + 1}}{{x - 1}}} \right)^\prime = \frac{{\left( {x + 1} \right)^\prime\left( {x - 1} \right) - \left( {x + 1} \right)\left( {x - 1} \right)^\prime}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{x - 1 - \left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{\cancel{x} - 1 - \cancel{x} + 1}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}.\]
At the point \({x_0} = 2,\) the function and the derivative have the following values:
\[y\left( 2 \right) = \frac{{2 + 1}}{{2 - 1}} = 3,\]
\[y^\prime\left( 2 \right) = f^\prime\left( 2 \right) = - \frac{2}{{{{\left( {2 - 1} \right)}^2}}} = - 2.\]
So the equation of the normal is given by
\[y - {y_0} = - \frac{1}{{f^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\]
\[y - 3 = - \frac{1}{{\left( { - 2} \right)}}\left( {x - 2} \right),\]
\[y - 3 = \frac{x}{2} - 1,\]
\[y = \frac{x}{2} + 2.\]
Example 10.
Find the equations of the tangent line and normal line to the parabola \[y = 2{x^2}\] at the point \(\left( {2,8} \right).\)
Solution.
Calculate the derivative of the function:
\[y^\prime = \left( {2{x^2}} \right)^\prime = 4x, \;\; \Rightarrow y^\prime\left( 2 \right) = 8.\]
Write the equation of the tangent line:
\[y - {y_0} = y^\prime\left( {{x_0}} \right)\left( {x - {x_0}} \right),\]
\[y - 8 = 8\left( {x - 2} \right),\]
\[y - 8 = 8x - 16,\]
\[8x - y - 8 = 0.\]
Similarly, we get the equation of the normal line:
\[y - {y_0} = - \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\]
\[y - 8 = - \frac{1}{8}\left( {x - 2} \right),\]
\[y - 8 = - \frac{x}{8} + \frac{1}{4},\]
\[8y - 64 = - x + 2,\]
\[x + 8y - 66 = 0.\]
So the answer is given by
\[8x - y - 8 = 0,\;\;x + 8y - 66 = 0.\]
Example 11.
Write an equation of the normal to the ellipse \[\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1\]
at the point \(\left( {1,\frac{{\sqrt 3 }}{2}} \right)\) (Figure \(5\)).
Solution.
Find the derivative \(y'\left( x \right)\) by implicit differentiation :
\[{\left( {\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1}} \right)^\prime } = 1',\;\; \Rightarrow \frac{{2x}}{4} + 2yy' = 0,\;\; \Rightarrow 4yy' = - x,\;\; \Rightarrow y^\prime = - \frac{x}{{4y}}.\]
The derivative is the point of tangency is equal to
\[y'\left( {{x_0},{y_0}} \right) = y'\left( {1,\frac{{\sqrt 3 }}{2}} \right) = - \frac{1}{{\frac{{4\sqrt 3 }}{2}}} = - \frac{1}{{2\sqrt 3 }}.\]
Then the equation of the normal is written as
\[y - {y_0} = - \frac{1}{{y'\left( {{x_0},{y_0}} \right)}}\left( {x - {x_0}} \right),\;\; \Rightarrow y - \frac{{\sqrt 3 }}{2} = - \frac{1}{{\left( { - \frac{1}{{2\sqrt 3 }}} \right)}}\left( {x - 1} \right),\;\; \Rightarrow y - \frac{{\sqrt 3 }}{2} = 2\sqrt 3 x - 2\sqrt 3 ,\;\; \Rightarrow y = 2\sqrt 3 x - 2\sqrt 3 + \frac{{\sqrt 3 }}{2},\;\; \Rightarrow y = 2\sqrt 3 x - \frac{{3\sqrt 3 }}{2} \approx 3,46x - 2,60.\]
Figure 5.
Example 12.
Find the angle at which the parabola \[y = {x^2} - \frac{1}{4}\] intersects the positive \(x-\)axis.
Solution.
The angle of intersection is equal to the slope angle \(\alpha\) of the tangent line.
Determine the point of tangency:
\[y = 0,\;\; \Rightarrow {x^2} - \frac{1}{4} = 0,\;\; \Rightarrow {x_0} = \frac{1}{2}.\]
Find the derivative at this point:
\[f^\prime\left( x \right) = \left( {{x^2} - 1} \right)' = 2x;\]
\[f^\prime\left( {{x_0}} \right) = f^\prime\left( {\frac{1}{2}} \right) = 1.\]
Since
\[\tan \alpha = f^\prime\left( {{x_0}} \right),\]
we get
\[\tan \alpha = 1,\;\; \Rightarrow \alpha = \arctan 1 = \frac{\pi }{4} = 45^\circ .\]
Example 13.
Find the angles at which the curve \[y = {x^3} - x\] intersects the \(x\)-axis.
Solution.
The cubic function \(y = {x^3} - x\) intersects the horizontal axis at the following points
\[f\left( x \right) = 0,\;\; \Rightarrow {x^3} - x = 0,\;\; \Rightarrow x\left( {x - 1} \right)\left( {x + 1} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_{2,3}} = \pm 1.\]
We calculate the values of the derivative at these points:
\[f'\left( x \right) = \left( {{x^3} - x} \right)^\prime = 3{x^2} - 1;\]
\[f'\left( 0 \right) = 3 \cdot {0^2} - 1 = - 1,\;\;f'\left( { - 1} \right) = 3 \cdot {\left( { - 1} \right)^2} - 1 = 2,\;\;f'\left( 1 \right) = 3 \cdot {1^2} - 1 = 2.\]
The angle at which the curve intersects the \(x\)-axis is determined by the angle of inclination of the tangent to the graph of the function at the point of intersection. In turn, the slope of the tangent is equal to the value of the derivative at the point of tangency. Consequently, we obtain the following values for the angles at points of intersection:
\({x_1} = 0,\;\;\) \(\Rightarrow f'\left( 0 \right) = - 1,\;\;\) \(\Rightarrow \tan {\alpha _1} = - 1,\;\;\) \(\Rightarrow {{\alpha _1} = {\frac{{3\pi }}{4}} = {135^{\circ};}}\)
\({x_2} = - 1,\;\;\) \(\Rightarrow{f'\left( { - 1} \right) = 2,\;\;}\) \(\Rightarrow {\tan {\alpha _2} = 2,\;\;}\) \(\Rightarrow {{\alpha _2} = \arctan 2 \approx 63^{\circ};}\)
\({x_3} = 1,\;\;\) \(\Rightarrow {f'\left( {1} \right) = 2,\;\;}\) \(\Rightarrow {\tan {\alpha _3} = 2,\;\;}\) \(\Rightarrow {{\alpha _3} = \arctan 2 \approx 63^{\circ}.}\)
Example 14.
The normal drawn to the hyperbola \[y = \frac{1}{x}\] at a point \({x_0}\) makes an angle of \(45^\circ\) with the positive \(x-\)axis. Determine the coordinate \({x_0}\) of the point of tangency (Figure \(6\)).
Solution.
Figure 6. It is known that the slope of the normal line is equal to \( - \frac{1}{{f'\left( {{x_0}} \right)}} .\) From the other side, it is equal to the tangent of the slope angle. Then we have the following relationship:
\[ - \frac{1}{{f^\prime\left( {{x_0}} \right)}} = \tan 45^\circ = 1.\]
As the derivative of the hyperbolic function is
\[f^\prime\left( x \right) = \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}},\]
we get
\[ - \frac{1}{{\left( { - \frac{1}{{x_0^2}}} \right)}} = 1,\]
or
\[x_0^2 = 1.\]
Taking into account that the tangency point is in the first quadrant, we find that
\[{x_0} = 1.\]
Example 15.
Write equations of the tangent and normal to the graph of the function \[y = x\sqrt {x - 1} \] at \(x = 2.\)
Solution.
Calculate the derivative for the given function:
\[y'\left( x \right) = \left( {x\sqrt {x - 1} } \right)^\prime = x'\sqrt {x - 1} + x{\left( {\sqrt {x - 1} } \right)^\prime } = \sqrt {x - 1} + \frac{x}{{2\sqrt {x - 1} }} = \frac{{2\left( {x - 1} \right) + x}}{{2\sqrt {x - 1} }} = \frac{{3x - 2}}{{2\sqrt {x - 1} }}.\]
At the point \(x = 2,\) the derivative is
\[y'\left( 2 \right) = \frac{{3 \cdot 2 - 2}}{{2\sqrt {2 - 1} }} = 2.\]
The value of the function at this point is
\[y\left( 2 \right) = 2 \cdot 1 = 2.\]
Find the equation of the tangent:
\[ y - {y_0} = y'\left( {{x_0}} \right)\left( {x - {x_0}} \right),\;\; \Rightarrow y - 2 = 2\left( {x - 2} \right),\;\; \Rightarrow y - 2 = 2x - 4,\;\; \Rightarrow y = 2x - 2,\]
and the equation of the normal at this point:
\[ y - {y_0} = - \frac{1}{{y'\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\;\; \Rightarrow y - 2 = - \frac{1}{2}\left( {x - 2} \right),\;\; \Rightarrow y - 2 = - \frac{x}{2} + 1,\;\; \Rightarrow y = - \frac{x}{2} + 3.\]
Example 16.
Find the equations of the tangent and normal lines to the graph of the natural logarithm function at \({x_0} = 1.\)
Solution.
Figure 7. Find the derivative
\[f^\prime\left( 1 \right) = \frac{1}{1} = 1.\]
Calculate \({y_0}:\)
\[{y_0} = \ln 1 = 0.\]
Then the equation of the tangent is
\[y - 0 = 1 \cdot \left( {x - 1} \right),\;\; \Rightarrow y = x - 1.\]
Respectively, the equation of the normal line is given by
\[y - 0 = - 1 \cdot \left( {x - 1} \right),\;\; \Rightarrow y = - x + 1.\]
Example 17.
Find the equation of the normal line to the curve \[y = \text{arccot}\frac{1}{x}\] at \(x = 1.\)
Solution.
First of all, we find the derivative. Using the chain rule, we obtain:
\[y^\prime = \left( \text{arccot} \frac{1}{x} \right)^\prime = - \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \cdot \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{1 + \frac{1}{{{x^2}}}}} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = \frac{{{x^2}}}{{1 + {x^2}}} \cdot \frac{1}{{{x^2}}} = \frac{1}{{1 + {x^2}}}.\]
Calculate the derivative value at \(x = 1:\)
\[y^\prime\left( 1 \right) = \frac{1}{{1 + {1^2}}} = \frac{1}{2}.\]
The value of the function itself is
\[{y_0} = y\left( 1 \right) = \text{arccot}{\,1} = \frac{\pi }{4}.\]
Now we have all data to write the equation of the normal.
\[y - {y_0} = - \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\]
\[y - \frac{\pi }{4} = - \frac{1}{{\frac{1}{2}}}\left( {x - 1} \right),\]
\[y - \frac{\pi }{4} = - 2x + 2,\]
\[8x + 4y - 8 - \pi = 0.\]
Example 18.
Given the parabola \[y = 2{x^2}.\] A secant is drawn through the \(2\) points of the parabola, which have the coordinates \(x = -1\) and \(x = 2\) (Figure \(8\)). Find the tangent to the parabola parallel to the secant.
Solution.
We first calculate the coordinates \(y\) of the chord \(KL\) (Figure \(8\)):
\[y\left( { - 1} \right) = 2 \cdot {\left( { - 1} \right)^2} = 2;\;\;y\left( 2 \right) = 2 \cdot {2^2} = 8.\]
Figure 8. Then the equation of the secant \(KL\) is written as
\[\frac{{y - {y_K}}}{{{y_L} - {y_K}}} = \frac{{x - {x_K}}}{{{x_L} - {x_K}}},\;\; \Rightarrow \frac{{y - 2}}{{8 - 2}} = \frac{{x - \left( { - 1} \right)}}{{2 - \left( { - 1} \right)}},\;\; \Rightarrow \frac{{y - 2}}{6} = \frac{{x + 1}}{3},\;\; \Rightarrow y - 2 = 2\left( {x + 1} \right),\;\; \Rightarrow y = 2x + 4,\]
that is, \(k = 2.\) The slope of the tangent has the same value \(k = 2.\)
Find the coordinates of the point of tangency from the condition \(y'\left( x \right) = k:\)
\[y'\left( x \right) = k,\;\; \Rightarrow {\left( {2{x^2}} \right)^\prime } = 2,\;\; \Rightarrow 4x = 2,\;\; \Rightarrow x = \frac{1}{2}.\]
Hence, the coordinate \(y\) of the point of tangency \(M\) is equal to
\[{y_M} = 2 \cdot {\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}.\]
Thus, the point of tangency \(M\) has the coordinates \(\left( {\frac{1}{2}, \frac{1}{2}} \right).\) From this we obtain the equation of the tangent in the following form:
\[y - {y_M} = k\left( {x - {x_M}} \right),\;\; \Rightarrow y - \frac{1}{2} = 2\left( {x - \frac{1}{2}} \right),\;\; \Rightarrow y - \frac{1}{2} = 2x - 1,\;\; \Rightarrow y = 2x - \frac{1}{2}.\]
Example 19.
A tangent line is drawn to the graph of the function \[y = \frac{1}{x}\] at the point \(\left( {1,1} \right)\) (Figure \(9\)). Find the length of the tangent segment \(AB\) in the first quadrant.
Solution.
Figure 9. First we find the derivative at the point of tangency:
\[f^\prime\left( x \right) = \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}},\]
\[f^\prime\left( 1 \right) = - 1.\]
Write the equation of the tangent line
\[y - 1 = - 1\left( {x - 1} \right),\]
\[y - 1 = - x + 1,\]
\[y = - x + 2.\]
Find at which points the tangent line intersects the coordinate axes:
\[A:\;x = 0,\;y = 2;\]
\[B:\;x = 2,\;y = 0.\]
As you can see, the tangent line forms the right triangle. So we can determine the length of the segment \(AB\) by the Pythagorean theorem :
\[AB = \sqrt {{2^2} + {2^2}} = \sqrt 8 .\]
Example 20.
The tangent and normal lines are drawn to the parabola \[y = {x^2}\] at the point \({x_0} = 2\) (Figure \(10\)). Find the length of the line segment \(AB\) between the points of intersection of the lines with the \(x-\)axis.
Solution.
Figure 10. We solve the problem in general form assuming \({x_0}\) is an arbitrary point.
The derivative of the quadratic function is
\[y^\prime = \left( {{x^2}} \right) = 2x.\]
At the point \({x_0},\) the function and the derivative take the values
\[{y_0} = x_0^2,\;\;y^\prime\left( {{x_0}} \right) = 2{x_0}.\]
Write the equation of the tangent line \(MA:\)
\[y - {y_0} = y^\prime\left( {{x_0}} \right)\left( {x - {x_0}} \right),\]
\[y - x_0^2 = 2{x_0}\left( {x - {x_0}} \right),\]
\[y - x_0^2 = 2{x_0}x - 2x_0^2,\]
\[y = 2{x_0}x - x_0^2.\]
The \(y-\)coordinate of the point \(A\) is zero. Then the \(x-\)coordinate of the point is equal to
\[2{x_0}{x_A} - x_0^2 = 0,\;\; \Rightarrow {x_A} = \frac{{{x_0}}}{2}.\]
Similarly we write the equation of the normal line passing through the point \(M:\)
\[y - {y_0} = - \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\]
\[y - x_0^2 = - \frac{1}{{2{x_0}}}\left( {x - {x_0}} \right),\]
\[y - x_0^2 = - \frac{x}{{2{x_0}}} + \frac{1}{2},\]
\[y = - \frac{x}{{2{x_0}}} + x_0^2 + \frac{1}{2}.\]
The point \(B\) has the coordinates \(B\left( {{x_B},0} \right).\) Then we find
\[ - \frac{{{x_B}}}{{2{x_0}}} + x_0^2 + \frac{1}{2} = 0,\]
\[{x_B} = 2x_0^3 + {x_0}.\]
Thus, we know the \(x-\)coordinates of the points \(A\) and \(B\). Then the length of the line segment \(AB\) is given by
\[AB = {x_B} - {x_A} = 2x_0^3 + {x_0} - \frac{{{x_0}}}{2} = 2x_0^3 + \frac{{{x_0}}}{2}.\]
Plug the value \({x_0} = 2\) in the last formula to get the numeric answer:
\[AB = 2x_0^3 + \frac{{{x_0}}}{2} = 2 \cdot {2^3} + \frac{2}{2} = 17.\]
