# Tangent and Normal Lines – Page 2

### Example 11.

Write an equation of the normal to the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ at the point $$\left( {1,\large\frac{{\sqrt 3 }}{2}\normalsize} \right)$$ (Figure $$5$$).

Solution.

Find the derivative $$y’\left( x \right)$$ by implicit differentiation:

${{\left( {\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1}} \right)^\prime } = 1′,\;\;}\Rightarrow {\frac{{2x}}{4} + 2yy’ = 0,\;\;}\Rightarrow {4yy’ = – x,\;\;}\Rightarrow {y = – \frac{x}{{4y}}.}$

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

${y – {y_0} = – \frac{1}{{y’\left( {{x_0},{y_0}} \right)}}\left( {x – {x_0}} \right),\;\;}\Rightarrow { y – \frac{{\sqrt 3 }}{2} = – \frac{1}{{\left( { – \frac{1}{{2\sqrt 3 }}} \right)}}\left( {x – 1} \right),\;\;}\Rightarrow {y – \frac{{\sqrt 3 }}{2} = 2\sqrt 3 x – 2\sqrt 3 ,\;\;}\Rightarrow {y = 2\sqrt 3 x – 2\sqrt 3 + \frac{{\sqrt 3 }}{2},\;\;}\Rightarrow {y = 2\sqrt 3 x – \frac{{3\sqrt 3 }}{2}} \approx {3,46x – 2,60.}$

### 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:

${f’\left( x \right) = {\left( {{x^3} – x} \right)^\prime } } = {3{x^2} – 1;}$

${f’\left( 0 \right) = 3 \cdot {0^2} – 1 = – 1,}\;\;\;\kern-0.3pt {f’\left( { – 1} \right) = 3 \cdot {\left( { – 1} \right)^2} – 1 = 2,}\;\;\;\kern-0.3pt {f’\left( 1 \right) = 3 \cdot {1^2} – 1 = 2.}$

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:

1. $${x_1} = 0,\;\;$$ $$\Rightarrow f’\left( 0 \right) = – 1,\;\;$$ $$\Rightarrow \tan {\alpha _1} = – 1,\;\;$$ $$\Rightarrow {{\alpha _1} = {\large\frac{{3\pi }}{4}\normalsize} = {135^{\circ};}}$$
2. $${x_2} = – 1,\;\;$$ $$\Rightarrow{f’\left( { – 1} \right) = 2,\;\;}$$ $$\Rightarrow {\tan {\alpha _2} = 2,\;\;}$$ $$\Rightarrow {{\alpha _2} = \arctan 2 \approx 63^{\circ};}$$
3. $${x_3} = 1,\;\;$$ $$\Rightarrow {f’\left( {1} \right) = 2,\;\;}$$ $$\Rightarrow {\tan {\alpha _3} = 2,\;\;}$$ $$\Rightarrow {{\alpha _3} = \arctan 2 \approx 63^{\circ}.}$$

### Example 14.

The normal drawn to the hyperbola $$y = \large{\frac{1}{x}}\normalsize$$ 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.

It is known that the slope of the normal line is equal to $$– \large{\frac{1}{{f’\left( {{x_0}} \right)}}}\normalsize .$$ 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.

Find the derivative

${f^\prime\left( x \right) = \left( {\ln x} \right)^\prime = \frac{1}{x}.}$

${f^\prime\left( 1 \right) = \frac{1}{1} = 1.}$

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 $$y = {\mathop{\rm arccot}\nolimits} \large{\frac{1}{x}}\normalsize$$ at $$x = 1.$$

Solution.

First of all, we find the derivative. Using the chain rule, we obtain:

${y^\prime = \left( {{\mathop{\rm arccot}\nolimits} \frac{1}{x}} \right)^\prime }={ – \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \cdot \left( {\frac{1}{x}} \right)^\prime }={ – \frac{1}{{1 + \frac{1}{{{x^2}}}}} \cdot \left( { – \frac{1}{{{x^2}}}} \right) }={ \frac{{{x^2}}}{{1 + {x^2}}} \cdot \frac{1}{{{x^2}}} }={ \frac{1}{{1 + {x^2}}}.}$

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

${y_0} = y\left( 1 \right) = {\mathop{\rm arccot}\nolimits} {\,1} = \frac{\pi }{4}.$

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 $$8$$):

${y\left( { – 1} \right) = 2 \cdot {\left( { – 1} \right)^2} = 2;}\;\;\;\kern-0.3pt {y\left( 2 \right) = 2 \cdot {2^2} = 8.}$

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 $$\left( {\large\frac{1}{2}\normalsize,\large\frac{1}{2}\normalsize} \right).$$ 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 $$y = \large{\frac{1}{x}}\normalsize$$ at the point $$\left( {1,1} \right)$$ (Figure $$9$$). Find the length of the tangent segment $$AB$$ in the first quadrant.

Solution.

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 Pythagoras’ 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.

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:

${AB = 2x_0^3 + \frac{{{x_0}}}{2} }={ 2 \cdot {2^3} + \frac{2}{2} }={ 17.}$

### Example 21.

Determine the area of the triangle formed by the tangent to the graph of the function $$y = 3 – {x^2}$$ drawn at the point $$\left( {1,2} \right)$$ and the coordinate axes (Figure $$11$$).

Solution.

Find the equation of the tangent. Given that

${f’\left( x \right) = \left( {3 – {x^2}} \right) = – 2x;\;\;}\Rightarrow {f’\left( 1 \right) = – 2,}$

we get the equation of the tangent in the following form:

${y – {y_M} = f’\left( {{x_M}} \right)\left( {x – {x_M}} \right),\;\;}\Rightarrow {y – 2 = – 2\left( {x – 1} \right),\;\;}\Rightarrow {y = – 2x + 4.}$

We transform it to the intercept form:

${y = – 2x + 4,\;\;}\Rightarrow { y + 2x = 4,\;\;}\Rightarrow {\frac{y}{4} + \frac{{2x}}{4} = 1,\;\;}\Rightarrow {\frac{y}{4} + \frac{x}{2} = 1.}$

It follows that the length of the segment $$OA$$ is $$4,$$ and the length of the segment $$OB$$ is $$2.$$ Then the area of the triangle $$OAB$$ is given by

$S = \frac{{\left| {OA} \right| \cdot \left| {OB} \right|}}{2} = \frac{{4 \cdot 2}}{2} = 4.$

### Example 22.

The normal drawn to the curve $$y = \ln x$$ at the point $$M\left( {1,0} \right)$$ intersects the $$y-$$axis at the point $$A$$ (Figure $$12$$). Find the area $$S$$ of the triangle $$AOM.$$

Solution.

The derivative of the natural logarithm is

$y^\prime = \left( {\ln x} \right)^\prime = \frac{1}{x}.$

At $${x_0} = 1,$$ we have

$y^\prime(1) = 1.$

Then the normal line at this point is defined by the equation

$y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),$

$y – 0 = – \frac{1}{1}\left( {x – 1} \right),$

or

$y = – x + 1.$

We see that the slope of the tangent line is equal to $$-1,$$ so the triangle $$AOM$$ is both right and isosceles. In this case, its area is

$S = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}.$

### Example 23.

A parabola is defined by the equation $$y = {x^2} + 2x + 3.$$ Write equations of the tangents to the parabola passing through the point $$A\left( { – 1,1} \right).$$

Solution.

We transform the equation of the parabola to the form

${y = {x^2} + 2x + 3 } = {{x^2} + 2x + 1 + 2 } = {{\left( {x + 1} \right)^2} + 2.}$

It can be seen that the graph of the parabola is obtained from the graph of the function $$y = {x^2}$$ by parallel shifting by 1 unit to the left and 2 units up (Figure $$13$$).

Let us find the equations of two tangents to the parabola passing through the point $$A\left( { – 1,1} \right).$$ Each of these tangents is defined by the equation

${y – {y_A} = k\left( {x – {x_A}} \right),\;\;}\Rightarrow { y – 1 = k\left( {x – \left( { – 1} \right)} \right),\;\;}\Rightarrow {y – 1 = kx + k,\;\;}\Rightarrow {y = kx + k + 1,}$

where $$k$$ is the slope ($${k_1}$$ for the first tangent and $${k_2}$$ for the second). Thus, the problem reduces to finding of the slopes of the tangents $${k_1}$$ and $${k_2}.$$ Take into account that at the points of tangency $$B$$ and $$C$$ the following condition holds:

${\left\{ \begin{array}{l} y = kx + k + 1\\ y = {x^2} + 2x + 3 \end{array} \right.,\;\;}\Rightarrow {kx + k + 1 = {x^2} + 2x + 3.}$

Also at the points of tangency $$B$$ and $$C,$$ the slope is equal to the derivative of the function $$y = {x^2} + 2x + 3.$$ Since

${y’ = {\left( {{x^2} + 2x + 3} \right)^\prime } } = {2x + 2,}$

then we obtain another equation in the form

$k = 2x + 2.$

As a result, we have the system of two equations

$\left\{ \begin{array}{l} kx + k + 1 = {x^2} + 2x + 3\\ k = 2x + 2 \end{array} \right.$

with two unknowns $$k$$ and $$x.$$ Solving this system, we find the values of $$k$$ and $$x$$ (i.e. the slopes of the tangents $${k_1},$$ $${k_2}$$ and $$x$$-coordinates of the points of tangency $$B$$ and $$C$$):

${\left\{ \begin{array}{l} kx + k + 1 = {x^2} + 2x + 3\\ k = 2x + 2 \end{array} \right.,\;\;}\Rightarrow {\left( {2x + 2} \right)x + 2x + 2 + 1} = {{x^2} + 2x + 3,\;\;}\Rightarrow {2{x^2} + 2x + 2x + 3 = {x^2} + 2x + 3,\;\;}\Rightarrow {{x^2} + 2x = 0,\;\;}\Rightarrow {{x_1} = – 2,\;{x_2} = 0.}$

The first solution $${x_1} = – 2$$ corresponds to point $$B.$$ The second solution $${x_2} = 0$$ is the coordinate of the point of tangency $$C.$$ The slopes have the following values:

1. tangent $$AB:\;$$ $${x_1} = -2,$$ $${k_1} = -2;$$
2. tangent $$AC:\;$$ $${x_2} = 0,$$ $${k_2} = 2.$$

Then the equations of the tangents to the parabola are given by

1. tangent $$AB:\;$$ $$y = -2x – 1;$$
2. tangent $$AC:\;$$ $$y = 2x + 3.$$

### Example 24.

Find the equation of the tangent line drawn to the curve $${y^4} – 4{x^4} – 6xy = 0$$ at the point $$M\left( {1,2} \right).$$

Solution.

This curve is given in implicit form. So we differentiate the equation implicitly to determine the derivative:

$\left( {{y^4} – 4{x^4} – 6xy} \right)^\prime = 0^\prime,$

$4{y^3}y^\prime – 16{x^3} – 6\left( {y + xy^\prime} \right) = 0,$

$4{y^3}y^\prime – 16{x^3} – 6y – 6xy^\prime = 0,$

$y^\prime\left( {2{y^3} – 3x} \right) = 3y + 8{x^3},$

$y^\prime = \frac{{3y + 8{x^3}}}{{2{y^3} – 3x}}.$

Calculate the value of the derivative at the point $$M\left( {1,2} \right):$$

$y^\prime(1,2) = \frac{{3 \cdot 2 + 8 \cdot {1^3}}}{{2 \cdot {2^3} – 3 \cdot 1}} = \frac{{14}}{{13}}.$

Now we can write the equation of the tangent line:

$y – 2 = \frac{{14}}{{13}}\left( {x – 1} \right),$

$y – 2 = \frac{{14}}{{13}}x – \frac{{14}}{{13}},$

$13y – 26 = 14x – 14,$

$14x – 13y + 12 = 0.$

### Example 25.

Find the equation of the normal line to the curve $${x^3} + {y^2} + 2x – 6 = 0$$ at the point $$\left( { – 1,3} \right).$$

Solution.

The given function is defined implicitly. We follow the same steps to get the equation of the normal as for an explicit function. First we find the derivative (using implicit differentiation):

$\left( {{x^3} + {y^2} + 2x – 6} \right)^\prime = 0^\prime,$

$3{x^2} + 2yy^\prime + 2 = 0,$

$2yy^\prime = – 3{x^2} – 2,$

$y^\prime = – \frac{{3{x^2} + 2}}{{2y}}.$

Substitute the coordinates of the tangency point:

${y^\prime = – \frac{{3 \cdot {{\left( { – 1} \right)}^2} + 2}}{{2 \cdot 3}} }={ – \frac{5}{6}.}$

Then we have the equation of the normal in the following form

$y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),$

$y – 3 = – \frac{1}{{\left( { – \frac{5}{6}} \right)}}\left( {x – \left( { – 1} \right)} \right),$

$y – 3 = \frac{6}{5}\left( {x + 1} \right),$

$y = \frac{6}{5}x + \frac{6}{5} + 3,$

$y = \frac{6}{5}x + \frac{{21}}{5},$

$6x – 5y + 21 = 0.$

### Example 26.

Prove that the curves $${x^2} – {y^2} = 3$$ and $$xy = 2$$ intersect at the right angle.

Solution.

These curves are hyperbolas shown schematically in Figure $$14.$$

Determine their points of intersection:

${\left\{ \begin{array}{l} {x^2} – {y^2} = 3\\ xy = 2 \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} {x^2} – {y^2} = 3\\ y = \frac{2}{x} \end{array} \right.,\;\;}\Rightarrow {{x^2} – {\left( {\frac{2}{x}} \right)^2} = 3,\;\;}\Rightarrow {\frac{{{x^2} – 4}}{{{x^2}}} = 3,\;\;}\Rightarrow {{x^4} – 3{x^2} – 4 = 0,\;\;}\Rightarrow {D = 9 + 16 = 25,\;\;}\Rightarrow {{x^2} = \frac{{3 \pm 5}}{2} = – 1;\;4.}$

Obviously, the points of intersection can be determined from the condition $${x^2} = 4:$$

${x^2} = 4,\;\; \Rightarrow {x_{1,2}} = \pm 2.$

Compute the $$y$$-coordinates of these points:

${{x_1} = – 2,\;\;}\Rightarrow {{y_1} = \frac{2}{{\left( { – 2} \right)}} = – 1;}$

${{x_2} = 2,\;\;}\Rightarrow {{y_2} = \frac{2}{2} = 1.}$

Now we find the derivatives of the given functions. We calculate the derivative of the first function using implicit differentiation:

${{x^2} – {y^2} = 3,\;\;}\Rightarrow {{\left( {{x^2} – {y^2}} \right)^\prime } = 3′,\;\;}\Rightarrow {2x – 2yy’ = 0,\;\;}\Rightarrow {y’ = \frac{x}{y}.}$

The derivative of the second function is written as

$y’ = {\left( {\frac{2}{x}} \right)^\prime } = – \frac{2}{{{x^2}}}.$

Determine the values of the derivatives at $$x = -2$$ (thus we find the slopes of the tangents to each hyperbola at this point) and make sure that the product of the slopes of the tangents at this point is equal to $$-1:$$

$\require{AMSmath.js} {{k_1} = {\left( {\frac{x}{y}} \right)_{\substack{ x = -2\\ y= -1}}} = \frac{{\left( { – 2} \right)}}{{\left( { – 1} \right)}} = 2;}\;\;\;\kern-0.3pt {{k_2} = {\left( { – \frac{2}{{{x^2}}}} \right)_{x = – 2}} = – \frac{1}{2};\;\;}\Rightarrow {{k_1}{k_2} = 2 \cdot \left( { – \frac{1}{2}} \right) = – 1.}$

We perform the same test for the second point of intersection:

${{k_1} = {\left( {\frac{x}{y}} \right)_{\substack{ x = 2\\ y= 1}}} = \frac{{2}}{{1}} = 2;\;\;}\kern0pt{{k_2} = {\left( { – \frac{2}{{{x^2}}}} \right)_{x = 2}} = – \frac{1}{2};\;\;}\Rightarrow {{k_1}{k_2} = 2 \cdot \left( { – \frac{1}{2}} \right) = – 1.}$

Thus, the curves intersect at the right angle at each of these points.

### Example 27.

At what points do the curves $$y = {x^2}$$ and $$y = \sqrt x$$ intersect? Find the angles of intersection between the curves at these points.

Solution.

Firstly, we determine the points of intersection of the two curves. Solve the following equation:

${{x^2} = \sqrt x , \;}\Rightarrow{ {x^4} = x, \;}\Rightarrow {{x^4} – x = 0,\; }\Rightarrow {x\left( {{x^3} – 1} \right) = 0.}$

The equation has two roots:

${x_1} = 0,\;{x_2} = 1.$

Find the angle between the curves at the first point $${x_1} = 0.$$ Calculate the values of the derivatives of both functions at this point:

${{f_1}^\prime\left( x \right) = {\left( {{x^2}} \right)^\prime } = 2x,\;}\Rightarrow{ {f_1}^\prime\left( {{x_1}} \right) = 2{x_1} = 0;}$

${{f_2}^\prime\left( x \right) = {\left( {\sqrt x } \right)^\prime } = \frac{1}{{2\sqrt x }},\;} \Rightarrow {{f_2}^\prime\left( {{x_1}} \right) = \frac{1}{{2\sqrt {{x_1}} }} = \frac{1}{{2 \cdot 0}} = \infty .}$

As the slope of the tangent line is equal to the derivative of the function at the point of tangency, then we immediately conclude that the tangent line to the first function is horizontal $$({\tan{\alpha _1} = 0,\;} \Rightarrow {{\alpha _1} = 0)},$$ and the tangent line to the second function is vertical $$\left({\tan{\alpha _2} = \infty ,\;} \Rightarrow {{\alpha _2} = \large{\frac{\pi }{2}}\normalsize}\right).$$ Therefore the angle $${\beta _1}$$ between the curves at the point $${x_1} = 0$$ is equal to

${{\beta _1} = {\alpha _2} – {\alpha _1} }={ \frac{\pi }{2} – 0 = \frac{\pi }{2}.}$

Now consider intersection of the curves at the second point $${x_2} = 1.$$ Similarly, calculate the derivatives:

${f_1}^\prime\left( {{x_2}} \right) = 2{x_2} = 2;$

${f_2}^\prime\left( {{x_2}} \right) = \frac{1}{{2\sqrt {{x_2}} }} = \frac{1}{2}.$

Further, we use the remarkable tangent subtraction formula

${\tan \left( {{\alpha _2} – {\alpha _1}} \right) }={ \frac{{\tan {\alpha _2} – \tan {\alpha _1}}}{{1 + \tan {\alpha _2}\tan {\alpha _1}}}.}$

Given that $$\tan {\alpha _1} = 2$$ and $$\tan {\alpha _2} = \large{\frac{1}{2}}\normalsize,$$ we have

${\tan{\beta _2} = \tan \left( {{\alpha _2} – {\alpha _1}} \right) }={ \frac{{\tan {\alpha _2} – \tan {\alpha _1}}}{{1 + \tan {\alpha _2}\tan {\alpha _1}}} }={ \frac{{\frac{1}{2} – 2}}{{1 + \frac{1}{2} \cdot 2}} }={ \frac{{ – \frac{3}{2}}}{2} }={ – \frac{3}{4}.}$

So, the angle between the curves at the second point of intersection is equal to

${{\beta _2} = \arctan \left( { – \frac{3}{4}} \right) }={ – \arctan \frac{3}{4}.}$

In the final answer, we indicate the absolute value of the angle, i.e. $${\beta _2} = \arctan \large{\frac{3}{4}}\normalsize.$$

We also compute the $$y-$$values of the points of intersection:

$y({x_1}) = x_1^2 = \sqrt {{x_1}} = 0;$

$y({x_2}) = x_2^2 = \sqrt {{x_2}} = 1.$

Thus, the final answer is given by

${\left( {0,0} \right):{\beta _1} = \frac{\pi }{2};\;\;} \kern0pt{\left( {1,1} \right):{\beta _2} = \arctan \frac{3}{4}.}$

### Example 28.

Find the distance between the origin and the normal to the curve $$y = {e^{2x}} + {x^2}$$ at the point where $$x = 0$$ (Figure $$16$$).

Solution.

Calculate the value of the function at the point $$x = 0:$$

$y\left( 0 \right) = {e^0} + {0^2} = 1.$

Find the derivative

${y^\prime = \left( {{e^{2x}} + {x^2}} \right)^\prime }={ 2{e^{2x}} + 2x.}$

Then the derivative value at $$x = 0$$ is

$y^\prime\left( 0 \right) = 2{e^0} + 2 \cdot 0 = 2.$

Write the equation of the normal passing through the point $$M\left( {0,1} \right):$$

$y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),$

$y – 1 = – \frac{1}{2}\left( {x – 0} \right),$

$y = – \frac{x}{2} + 1.$

We find from the equation that

${ON} = 2,\; {OM} = 1.$

The distance $$L$$ between the origin and the normal line is the line segment $$OP$$, which is the altitude of the right triangle $$OMN.$$ The altitude $$L=OP$$ is given by the formula

$L = OP = \frac{{ON \cdot OM}}{{MN}}.$

Using the Pythagorean theorem, we have

$MN = \sqrt {O{N^2} + O{M^2}}.$

Thus, the distance $$L$$ is equal to

${L = OP = \frac{{ON \cdot OM}}{{MN}} }={ \frac{{ON \cdot OM}}{{\sqrt {O{N^2} + O{M^2}} }} }={ \frac{{2 \cdot 1}}{{\sqrt {{2^2} + {1^2}} }} }={ \frac{2}{{\sqrt 5 }}.}$

### Example 29.

Find the angle between the tangent to the cardioid $$r = a\left( {1 + \cos \theta } \right)$$ and the radius vector of the point of tangency (Figure $$17$$).

Solution.

The given angle (Figure $$17$$) is calculated by the formula

$\tan \omega = \frac{r}{{{r’_\theta }}}.$

Here

${{r’_\theta } = {\left[ {a\left( {1 + \cos \theta } \right)} \right]^\prime } } = { – a\sin \theta .}$

Consequently,

$\require{cancel} {\tan \omega = \frac{{a\left( {1 + \cos \theta } \right)}}{{\left( { – a\sin \theta } \right)}} } = { – \frac{{1 + \cos \theta }}{{\sin\theta }} } = { – \frac{{\cancel{2}{{\cos }^{\cancel{2}}}\frac{\theta }{2}}}{{\cancel{2}\sin \frac{\theta }{2}\cancel{\cos \frac{\theta }{2}}}} } = { – \cot \frac{\theta }{2} } = {\tan \left( {\frac{\theta }{2} + \frac{\pi }{2}} \right).}$

In the last expression we have used a reduction identity. Thus, the angle between the tangent and the radius vector is

$\omega = \frac{\theta }{2} + \frac{\pi }{2}.$

### Example 30.

Find equation of the tangent and normal to the astroid $$x = a\,{\cos ^3}t,$$ $$y = a\,{\sin ^3}t$$ at the point $$t = \large\frac{\pi }{4}\normalsize$$ (Figure $$18$$).

Solution.

Calculate the derivatives of the parametric function:

${{x’_t} = {\left( {a\,{{\cos }^3}t} \right)^\prime } = – 3a\,{\cos ^2}t\sin t;}\;\;\;\kern-0.3pt {{y’_t} = {\left( {a\,{{\sin }^3}t} \right)^\prime } = 3a\,{\sin ^2}t\cos t.}$

Consequently,

${{y’_x} = \frac{{{y’_t}}}{{{x’_t}}} } = {\frac{{3a\,{{\sin }^2}t\cos t}}{{\left( { – 3a\,{{\cos }^2}t\sin t} \right)}} } = { – \frac{{\sin t}}{{\cos t}} } = { – \tan t.}$

By the reduction formula, we can write

$– \tan t = \tan \left( {\pi – t} \right).$

Since $$\tan \alpha = {y’_x} = \tan \left( {\pi – t} \right),$$ the angle $$\alpha$$ is equal to

${\alpha = \pi – t } = {\pi – \frac{\pi }{4} } = {\frac{{3\pi }}{4} = 135^{\circ}.}$

Then the derivative of the cardioid and, accordingly, the slope of the tangent at the point of tangency are equal

${y’_x}\left( {\frac{\pi }{4}} \right) = \tan \frac{{3\pi }}{4} = – 1.$

Find the coordinates of the point of tangency:

${{x_0} = x\left( {\frac{\pi }{4}} \right) } = {a\,{\cos ^3}\frac{\pi }{4} } = {a{\left( {\frac{{\sqrt 2 }}{2}} \right)^3} } = {\frac{{a\sqrt 2 }}{4},}$

${{y_0} = y\left( {\frac{\pi }{4}} \right) } = {a\,{\sin ^3}\frac{\pi }{4} } = {a{\left( {\frac{{\sqrt 2 }}{2}} \right)^3} } = {\frac{{a\sqrt 2 }}{4}.}$

Now we can write the equation of the tangent:

${y – {y_0} = {y’_x}\left( {{x_0}} \right)\left( {x – {x_0}} \right),\;\;}\Rightarrow {y – \frac{{a\sqrt 2 }}{4} = – 1\left( {x – \frac{{a\sqrt 2 }}{4}} \right),\;\;}\Rightarrow {y – \frac{{a\sqrt 2 }}{4} = – x + \frac{{a\sqrt 2 }}{4},\;\;}\Rightarrow {y = – x + \frac{{a\sqrt 2 }}{2}}$

and the equation of the normal:

${y – {y_0} = – \frac{1}{{{{y’}_x}\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),\;\;}\Rightarrow {y – \frac{{a\sqrt 2 }}{4} = – \frac{1}{{\left( { – 1} \right)}}\left( {x – \frac{{a\sqrt 2 }}{4}} \right),\;\;}\Rightarrow{y – \cancel{\frac{{a\sqrt 2 }}{4}} = x – \cancel{\frac{{a\sqrt 2 }}{4}},\;\;}\Rightarrow{y = x.}$

### Example 31.

A tangent is drawn to the graph of the function $$y =\cos x$$ at the point $$M\left( {{x_0},{y_0}} \right),$$ where $$0 \lt {x_0} \lt \large\frac{\pi }{2}\normalsize$$ (Figure $$19$$). Find the value of $${x_0},$$ at which the area of the triangle formed by the tangent and the coordinate axes will be the least.

Solution.

As the derivative of the cosine is

$y’\left( x \right) = {\left( {\cos x} \right)^\prime } = – \sin x,$

the slope of the tangent is given by

$\tan \alpha = – \sin {x_0} = y’\left( {{x_0}} \right).$

Then the equation of the tangent has the form

${y – {y_0} = y’\left( {{x_0}} \right)\left( {x – {x_0}} \right),\;\;}\Rightarrow {y – \cos {x_0} = – \sin {x_0}\left( {x – {x_0}} \right),\;\;}\Rightarrow {y – \cos {x_0} }={ \left( { – \sin {x_0}} \right)x – \left( { – \sin {x_0}} \right){x_0},\;\;}\Rightarrow {y + \left( {\sin {x_0}} \right)x = \cos {x_0} + \left( {\sin {x_0}} \right){x_0}}$

We represent it in the intercept form:

$\frac{x}{p} + \frac{y}{q} = 1.$

Hence,

${\frac{y}{{\cos {x_0} + \left( {\sin {x_0}} \right){x_0}}} }\kern0pt {+ \frac{{\left( {\sin {x_0}} \right)x}}{{\cos {x_0} + \left( {\sin {x_0}} \right){x_0}}} = 1.}$

that is the legs of the right triangle $$OAB$$ are

${\left| {OA} \right| = q = \cos {x_0} + \left( {\sin {x_0}} \right){x_0},}\;\;\;\kern-0.3pt {\left| {OB} \right| = p = \frac{{\cos {x_0} + \left( {\sin {x_0}} \right){x_0}}}{{\sin {x_0}}}.}$

For convenience, we denote $${x_0} = z$$ and express the area of the triangle $$OAB$$ as a function $$S\left( z \right):$$

${S = S\left( z \right) = \frac{{pq}}{2} } = {\frac{{{{\left( {\cos z + z\sin z} \right)}^2}}}{{2\sin z}}.}$

Investigate the extreme values of the function $$S\left( z \right).$$ Its derivative is given by

${S’\left( z \right) = \frac{1}{2}{\left[ {\frac{{{{\left( {\cos z + z\sin z} \right)}^2}}}{{\sin z}}} \right]^\prime } } = {\frac{{\left( {\cos z + z\sin z} \right)}}{{2\,{{\sin }^2}z}}} \cdot {\left[ {{z\cos z\sin z} – {{\cos }^2}z} \right].}$

Since in the interval $$0 \lt z \lt {\large\frac{\pi }{2}\normalsize}$$

$\cos z + z\sin z \gt 0,$

then the derivative has only one critical point that is determined by the condition

${z\cos z\sin z – {\cos ^2}z = 0,\;\;}\Rightarrow {\cos z\left( {z\sin z – \cos z} \right) = 0,\;\;}\Rightarrow {z – \cot z = 0.}$

This equation is solved numerically. However, it can be seen that if $$z = {\large\frac{\pi }{4}\normalsize},$$ then the left-hand side is negative:

${z = \frac{\pi }{4}:}\;\;\;\kern-0.3pt {z – \cot z = \frac{\pi }{4} – \cot \frac{\pi }{4} } = {\frac{\pi }{4} – 1 \approx – 0,21 \lt 0.}$

and at $$z = {\large\frac{\pi }{3}\normalsize},$$ the left-hand side is positive:

${z = \frac{\pi }{3}:}\;\;\;\kern-0.3pt {z – \cot z = \frac{\pi }{3} – \cot \frac{\pi }{3} } = {\frac{\pi }{3} – \frac{1}{{\sqrt 3 }} \approx 0,47 \gt 0.}$

Consequently, an extreme point of the function $$S\left( z \right)$$ is in the range of angles $$\left( {\large\frac{\pi }{4}\normalsize,\large\frac{\pi }{3}\normalsize} \right)$$ (Figure $$20$$) and this point is a point of minimum (judging by the change of sign of the derivative).

The approximate coordinate of the point of minimum is about $$0.86\;\text{rad}$$ or $$49,3^{\circ}.$$

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Problems 1-10
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Problems 11-31