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Precalculus Study Guide

Chapter 11: Functions Defined Implicitly

Introduction

Some equations of the form $f (x, y) = 0$ can be rewritten in the form $y = g (x)$ , in which case, we say that the original equation defined $y (x)$ implicitly, and the equation $y = g (x)$ gives $y (x)$ explicitly.

Some equations of the form $f (x, y) = 0$ cannot be rewritten in the form $y = g (x)$ . For these, there is no combination of algebraic steps that will isolate the letter $y$ in the equation. However, such equations can still define $y = g (x)$ in principle, even if the values have to be computed numerically. Such functions $y (x)$ are said to be defined implicitly by the equation $f (x, y) = 0$ .

Some equations of the form $f (x, y) = 0$ do not define $y (x)$ implicitly since they represent relations, one-many maps from the domain to the range. However, in such cases, we can define branches along which a function $y (x)$ is defined implicitly by the equation $f (x, y) = 0$ .

This chapter explores examples of all three types of behaviors, the simple case such as posed by the equation

$2 x + 3 y - 7 = 0$

where $y$ can easily be isolated; the single-branch case such as posed by the equation

$\ln (y) + \sqrt{x + y} - 5 = 0$

where $y$ cannot be isolated; and the multiple-branch case such as posed by

$x^{2} + y^{2} = 1$

the familiar equation of the unit circle.

Chapter Glossary

The following terms in Chapter 11 are linked to the Maple Math Dictionary.

discretization

discriminant

domain

explicit

floating-point

implicit

inequality

interval

nonnegative

parabola

quadratic equation
quadratic formula

real number

sequence

square root

Typical Problems

11.1. The equation $5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$ defines one or more functions $y = y (x)$ implicitly.

(a) Obtain a graph of the implicitly defined functions.

(b) Express explicitly each function in the graph plotted in Part (a).

11.2. The equation $\ln (x e^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$ implicitly defines a function $y = y (x)$ .

(a) On the interval $[1, 5]$ , obtain (using some sort of plotting device) a graph of the implicitly defined function.

(b)	In the interval $[2, 5]$ , substitute each node $x_{k} = 2 + (\frac{3}{20}) k, k = 0, 1, .. &period;, 20$ , into the given equation, then solve for the corresponding $y (x_{k})$ numerically. Plot the points $(x_{k}, y (x_{k}))$ and compare the graph to the one obtained in Part (a).

Maple Initializations

Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.

Solutions

Problem 11.1

11.1 - Mathematical Solution

11.1 (a) - Mathematical Solution

Obtaining a graph of the function defined implicitly by the equation

$5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$

is a challenging task if done without the aid of an appropriate plotting device. One way to accomplish this by hand is to anticipate Part (b) and obtain explicit expressions for the two branches contained therein. Thus, write the equation in the form

$2 y^{2} + (3 x - 7) y + (5 x^{2} - 4 x - 9) = 0$

and use the quadratic formula to obtain

$y_{\pm}$	$= \frac{- (3 x - 7) \pm \sqrt{{(3 x - 7)}^{2} - 4 (2) (5 x^{2} - 4 x - 9)}}{2 (2)}$
	$= \frac{1}{4} (7 - 3 x \pm \sqrt{121 - 10 x - 31 x^{2}})$

The square roots appearing in each of these branches is real for $121 - 10 x - 31 x^{2} \geq 0$ . Again applying the quadratic formula and sketching the parabola represented by the left-hand side of this inequality, we discover that the domain for each of the two branches of the implicit function is

$- \frac{5 + 8 \sqrt{59}}{31} \leq x \leq \frac{- 5 + 8 \sqrt{59}}{31}$

or approximately

$- 2.14 \leq x \leq 1.82$

Then, from the function values listed in Table 11.1.1, use a sheet of graph paper to sketch a graph akin to the one in Figure 11.1.1.

$\begin{array}{c} x & y_{+} (x) & y_{-} (x) \\ - 2.0 & 4.28 & 2.22 \\ - 1.8 & 4.65 & 1.55 \\ - 1.6 & 4.85 & 1.05 \\ - 1.4 & 4.95 & 0.65 \\ - 1.2 & 5.00 & 0.30 \\ - 1.0 & 5.00 & 0.00 \\ - 0.8 & 4.96 & - 0.26 \\ - 0.6 & 4.89 & - 0.49 \\ - 0.4 & 4.79 & - 0.69 \\ - 0.2 & 4.66 & - 0.86 \end{array}$	$\begin{array}{c} x & y_{+} (x) & y_{-} (x) \\ 0.0 & 4.50 & - 1.00 \\ 0.2 & 4.31 & - 1.11 \\ 0.4 & 4.10 & - 1.20 \\ 0.6 & 3.85 & - 1.25 \\ 0.8 & 3.56 & - 1.26 \\ 1.0 & 3.24 & - 1.24 \\ 1.2 & 2.86 & - 1.16 \\ 1.4 & 2.40 & - 1.00 \\ 1.6 & 1.82 & - 0.72 \\ 1.8 & 0.80 & 0.00 \end{array}$
Table 11.1.1 Function values for branches $y_{+}$ and $y_{-}$		Figure 11.1.1 Graph of implicitly-defined $y (x)$

11.1 (b) - Mathematical Solution

In Part (a), it was necessary to anticipate Part (b) where the two branches contained implicitly in the equation

$5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$

are obtained explicitly. Recall that by use of the quadratic formula, applied to

$2 y^{2} + (3 x - 7) y + (5 x^{2} - 4 x - 9) = 0$

the branches $y_{+}$ and $y_{-}$ were determined to be

$y_{\pm}$	$= \frac{- (3 x - 7) \pm \sqrt{{(3 x - 7)}^{2} - 4 (2) (5 x^{2} - 4 x - 9)}}{2 (2)}$
	$= \frac{1}{4} (7 - 3 x \pm \sqrt{121 - 10 x - 31 x^{2}})$

11.1 (c) - Mathematical Solution

If the substitution $x = 1$ is made in the equation

$5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$

the equation

$2 y^{2} - 4 y - 8$ = 0

results. This equation determines the two values of $y$ that correspond to $x = 1$ . Of course, this equation can be solved by the quadratic formula, yielding

$y_{\pm}$	$= \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (2) (- 8)}}{2 (2)}$
	$= \frac{4 \pm \sqrt{80}}{4}$
	$= 1 \pm \sqrt{5}$
	$= {\begin{array}{c} 3.236067977 \\ - 1.236067977 \end{array}$

11.1 - Maplet Solution

11.1 (a) - Maplet Solution

A graph of the implicit functions defined by the equation

$5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$

can be obtained with the Implicit Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 11.1.2.

For the graph, simply enter the defining equation and click the button labeled Graph. Use the Plot Options button as needed.

To launch the Implicit Function Tutor, click the following link: Implicit Function Tutor

Figure 11.1.2 Thumbnail image of the Implicit Function Tutor

11.1 (b) - Maplet Solution

The explicit rules for the functions defined implicitly by the equation

$5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$

can be obtained with the Implicit Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image Figure 11.1.2.

After entering the defining equation, click the button labeled Explicit Branches. The resulting expressions are then

$y_{\pm} = \frac{1}{4} (7 - 3 x \pm \sqrt{121 - 10 x - 31 x^{2}})$

Figure 11.1.2 Thumbnail image of the Implicit Function Tutor

To launch the Implicit Function Tutor, click the following link: Implicit Function Tutor

11.1 (c) - Maplet Solution

To use the given equation to compute all real values of $y (1)$ , again use the Implicit Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 11.1.3.

In the section Numeric study, enter 1 as the value of $x$ , and for the Interval containing corresponding $y$ , enter $y = - 2$ to 5, as suggested by the graph.

Click the button labeled Equation for y to see the equation that results from substituting $x = 1$ into the original equation.

Click the button labeled Analytic Solution to obtain the two values $y = 1 \pm \sqrt{5}$ or click the button labeled Numeric Solution to obtain the values of $y$ in floating-point (decimal) form.

Figure 11.1.3 Thumbnail image of the Implicit Function Tutor

To launch the Implicit Function Tutor, click the following link: Implicit Function Tutor

11.1 - Interactive Solution

11.1 (a) - Interactive Solution

•	Type (or Control-drag) the equation.

•	Context Panel: Plots≻2-D Implicit Plot≻ $x, y$

11.1 (b) - Interactive Solution

•	Type (or Control-drag) the equation.

•	Context Panel: Solve≻Solve for Variable≻ $y$

11.1 (c) - Interactive Solution

•	Type (or Control-drag) the equation. Press the Enter key.

•	Context Panel: Evaluate at a Point≻ $x = 1$

•	Context Panel: Solve

11.1 - Programmatic Solution

11.1 (a) - Programmatic Solution

The function defined implicitly by the equation

>	$q ≔ 5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$

can be plotted with Maple's implicitplot command found in the plots package. Its use leads to Figure 11.1.4.

>	$implicitplot (q, x = - 3 .. 2, y = - 2 .. 5, scaling = constrained)$

Figure 11.1.4 Graph of the function defined implicitly by $5 x^{2} + 3 x y + 2 y^{2} - 4 x - 7 y - 9 = 0$

11.1 (b) - Programmatic Solution

The equation

$q$

can be solved explicitly for $y = y (x)$ . Maple's solve command gives

>	$Y ≔ solve (q, y)$

Thus, there are two branches, which we label as

>	$y_{+} = Y [1] &semi; y_{-} = Y [2]$

These two explicit branches of the implicitly defined function are plotted in Figure 11.1.5, where $y_{+}$ appears in black, and $y_{-}$ appears in red.

>	$plot ([Y], x = - 2.2 .. 2, color = [black, red], numpoints = 2000, scaling = constrained, legend = [typeset (y_{+}), typeset (y_{-})])$

Figure 11.1.5 Graph of branches $y_{+}$ (black) and $y_{-}$ (red)

The algebra by which these two explicit branches are extracted from the equation

$q$

is no more than an application of the quadratic formula. The variable in the quadratic equation is $y$ , and all $x$ 's are treated as constants. Thus, the equation is written as

$A y^{2} + B y + C = 0$

where $A, B$ , and $C$ , are respectively

>	$A ≔ coeff (lhs (q), y, 2) &semi; B ≔ coeff (lhs (q), y, 1) &semi; C ≔ coeff (lhs (q), y, 0)$

The quadratic formula gives the two solutions for $y$ as

>	$(- B + sqrt (B^{2} - 4 A C)) / 2 / A &semi; (- B - sqrt (B^{2} - 4 A C)) / 2 / A$

in agreement with Maple's earlier solution.

The discriminant of this quadratic in the variable $y$ is

>	$simplify (B^{2} - 4 A C)$

which can also be obtained with Maple's discrim command, as shown by

>	$d ≔ discrim (lhs (q), y)$

This discriminant is nonnegative for $x$ in the interval

>	$r ≔ solve (d \geq 0, \{x\})$

or, in floating-point form,

>	$evalf (r)$

These calculations are consistent with Figure 11.1.5.

11.1 (c) - Programmatic Solution

To compute the value(s) of $y (1)$ , substitute $x = 1$ into the given equation, namely, into

$q$

obtaining

>	$q1 ≔ eval (q, x = 1)$

which is a quadratic equation in the variable $y$ . The solution can be obtained with the quadratic formula, or with Maple's solve command. In either event, the solutions will be

>	$solve (q1, y)$

a result consistent with evaluating the branches

>	$y [1] = Y [1] &semi; y [2] = Y [2]$

at $x = 1$ . Indeed, we find

>	$simplify (eval (Y [1], x = 1)) &semi; simplify (eval (Y [2], x = 1))$

Problem 11.2

11.2 - Mathematical Solution

11.2 (a) - Mathematical Solution

Figure 11.2.1 contains a (Maple-generated) graph of the function $y (x)$ defined implicitly by the equation

$\ln (x e^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$

Figure 11.2.1 Graph of $y (x)$ defined implicitly by the equation $\ln (x e^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$

11.2 (b) - Mathematical Solution

At a node $x_{k} = 2 + (\frac{3}{20}) k, k = 0, 1, ..., 20$ , the equation

$\ln (x e^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$

becomes one of the entries in Table 11.2.1.

$k$	Equation	$k$	Equation
0	$\ln (2) - 4 y + \sqrt{2 + y} + 15 = 0$	11	$\ln (\frac{73}{20}) + \frac{399}{20} - \frac{49}{4} y + \frac{1}{10} \sqrt{365 + 100 y} = 0$
1	$\ln (\frac{43}{20}) + \frac{309}{20} - \frac{19}{4} y + \frac{1}{10} \sqrt{215 + 100 y} = 0$	12	$\ln (\frac{19}{5}) - 13 y + \frac{1}{5} \sqrt{95 + 25 y} + \frac{102}{5} = 0$
2	$\ln (\frac{23}{10}) - \frac{11}{2} y + \frac{1}{10} \sqrt{230 + 100 y} + \frac{159}{10} = 0$	13	$\ln (\frac{79}{20}) + \frac{417}{20} - \frac{55}{4} y + \frac{1}{10} \sqrt{395 + 100 y} = 0$
3	$\ln (\frac{49}{20}) + \frac{327}{20} - \frac{25}{4} y + \frac{1}{10} \sqrt{245 + 100 y} = 0$	14	$\ln (\frac{41}{10}) - \frac{29}{2} y + \frac{1}{10} \sqrt{410 + 100 y} + \frac{213}{10} = 0$
4	$\ln (\frac{13}{5}) - 7 y + \frac{1}{5} \sqrt{65 + 25 y} + \frac{84}{5} = 0$	15	$\ln (\frac{17}{4}) + \frac{87}{4} - \frac{61}{4} y + \frac{1}{2} \sqrt{17 + 4 y} = 0$
5	$\ln (\frac{11}{4}) + \frac{69}{4} - \frac{31}{4} y + \frac{1}{2} \sqrt{11 + 4 y} = 0$	16	$\ln (\frac{22}{5}) + \frac{111}{5} - 16 y + \frac{1}{5} \sqrt{110 + 25 y} = 0$
6	$\ln (\frac{29}{10}) - \frac{17}{2} y + \frac{1}{10} \sqrt{290 + 100 y} + \frac{177}{10} = 0$	17	$\ln (\frac{91}{20}) + \frac{453}{20} - \frac{67}{4} y + \frac{1}{10} \sqrt{455 + 100 y} = 0$
7	$\ln (\frac{61}{20}) + \frac{363}{20} - \frac{37}{4} y + \frac{1}{10} \sqrt{305 + 100 y} = 0$	18	$\ln (\frac{47}{10}) - \frac{35}{2} y + \frac{1}{10} \sqrt{470 + 100 y} + \frac{231}{10} = 0$
8	$\ln (\frac{16}{5}) - 10 y + \frac{1}{5} \sqrt{80 + 25 y} + \frac{93}{5} = 0$	19	$\ln (\frac{97}{20}) + \frac{471}{20} - \frac{73}{4} y + \frac{1}{10} \sqrt{485 + 100 y} = 0$
9	$\ln (\frac{67}{20}) + \frac{381}{20} - \frac{43}{4} y + \frac{1}{10} \sqrt{335 + 100 y} = 0$	20	$\ln (5) - 19 y + \sqrt{5 + y} + 24 = 0$
10	$\ln (\frac{7}{2}) - \frac{23}{2} y + \frac{1}{2} \sqrt{14 + 4 y} + \frac{39}{2} = 0$
Table 11.2.1 The equation $\ln (x e^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$ at the nodes $x_{k} = 2 + (\frac{3}{20}) k, k = 0, 1, ..., 20$

The solutions of the equations in Table 11.2.1 appear in Table 11.2.2. Figure 11.2.2 contains a graph of the data points in Table 11.2.2.

$\begin{array}{c} x & y (x) \\ 2.00 & 4.56 \\ 2.15 & 3.93 \\ 2.30 & 3.48 \\ 2.45 & 3.14 \\ 2.60 & 2.87 \\ 2.75 & 2.66 \\ 2.90 & 2.48 \end{array}$	$\begin{array}{c} x & y (x) \\ 3.05 & 2.33 \\ 3.20 & 2.21 \\ 3.35 & 2.10 \\ 3.50 & 2.01 \\ 3.65 & 1.93 \\ 3.80 & 1.85 \\ 3.95 & 1.79 \end{array}$	$\begin{array}{c} x & y (x) \\ 4.10 & 1.73 \\ 4.25 & 1.68 \\ 4.40 & 1.63 \\ 4.55 & 1.59 \\ 4.70 & 1.55 \\ 4.85 & 1.52 \\ 5.00 & 1.48 \end{array}$
Table 11.2.2 Solutions of equations in Table 11.2.1

Figure 11.2.2 The curve from Figure 11.2.1 and the points from Table 11.2.2

11.2 - Maplet Solution

11.2 (a) - Maplet Solution

The function $y (x)$ defined implicitly by the equation

$\ln (x {&ExponentialE;}^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$

can be graphed on an interval such as $[1, 5]$ by means of the Implicit Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 11.2.3.

After entering the defining equation, use the Plot Option button to set the plotting interval.

Click the Graph button to obtain the required graph.

Figure 11.2.3 Thumbnail image of the Implicit Function Tutor

To launch the Implicit Function Tutor, click the following link: Implicit Function Tutor

11.2 (b) - Maplet Solution

The function $y (x)$ defined implicitly by the equation

$\ln (x {&ExponentialE;}^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$

can be graphed pointwise on an interval such as $[2, 5]$ by means of the Implicit Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 12.2.4.

For each of the nodes $x_{k} = 2 + (\frac{3}{20}) k, k = 0, 1, ..., 20$ , enter a single value of $x$ and click the button labeled Numeric Solution to obtain the corresponding value of $y$ .

Figure 11.2.4 Thumbnail image of the Implicit Function Tutor

One such calculation for $x = 2$ is shown in the solution provided where the nonlinear equation defining this value of $y$ has also been obtained by clicking the button labeled Equation for $y$ .

After collecting the coordinates of all 21 points, plot the points, connecting them with a smooth curve. Of course, this task is tedious if executed by hand. For this problem, the sections Interactive Solution and Programmatic Solution implement more efficient strategies for these calculations.

To launch the Implicit Function Tutor, click the following link: Implicit Function Tutor

11.2 - Interactive Solution

11.2 (a) - Interactive Solution

•	Type (or Control-drag) the equation, being sure to use the exponential "e" and not just the letter e.

•	Context Panel: Plots≻2-D Implicit Plot≻ $x, y$

11.2 (b) - Interactive Solution

Write the left-hand side of the equation as a function $f (x)$

•	Type (or Control-drag) the equation, being sure to use the exponential "e" and not just the letter e.

•	Press the Enter key.

•	Context Panel: Left-hand Side

•	Context Panel: Conversions≻Operator≻ $x$

•	Context Panel: Assign to a Name≻ $f$

Generate a list of $y$ -values

•	Type $f (2 + \frac{3}{20} k) = 0$ and press the Enter key.

•	Context Panel: Sequence≻ $k$ ≻ $0 .. 20$

•	Context Panel: Map Command Onto Map the command: fsolve Append the arguments: $y$

•	Context Panel: Conversions≻To List

•	Context Panel: Assign to a Name≻ $Y$

Generate corresponding list of $x$ -values

•	Type $2 + \frac{3}{20} k$ and press the Enter key.

•	Context Panel: Sequence≻ $k$ ≻ $0 .. 20$

•	Context Panel: Conversions≻To List

•	Context Panel: Assign to a Name≻ $X$

Graph computed points

•	Type $X, Y$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder≻Plot

Graph implicit function

•	Type (or Control-drag) the equation, being sure to use the exponential "e" and not just the letter e.

•	Context Panel: Plots≻Plot Builder $2 \leq x \leq 5$

Control-drag (or copy/paste) the graph of the implicit function onto the graph of the computed points.

11.2 - Programmatic Solution

11.2 (a) - Programmatic Solution

The function $y (x)$ defined implicitly by the equation

>	$q ≔ \ln (x \exp (- y)) + sqrt (x + y) - 5 x y + 3 x + 7 y + 9 = 0$

can be plotted with Maple's implicitplot command from the plots package. The result is seen in Figure 11.2.5.

>	$implicitplot (q, x = 1 .. 5, y = 1 .. 5)$

Figure 11.2.5 Graph of $y (x)$ defined implicitly by the equation $\ln (x e^{- y}) + \sqrt{x + y} - 5 x y + 3 x + 7 y + 9 = 0$

11.2 (b) - Programmatic Solution

The suggested discretization contains the nodes $x_{0} = 2$ and $x_{20} = 5$ . At the node $x_{2}$ , the equation

$q$

becomes

>	$q0 ≔ eval (q, x = 2)$

This equation in $y$ can be solved numerically, and for this, we use Maple's fsolve command. The result is

>	$y0 ≔ fsolve (q0, y)$

Thus, one point on the graph of the implicitly defined function $y = y (x)$ is

>	$`` (2, y0)$

At the other end of the interval $[2, 5]$ , we have $x_{20} = 5$ , and the calculations at that node are

>	$q20 ≔ eval (q, x = 5)$

and

>	$y20 ≔ fsolve (q20, y)$

Thus, the point

>	$`` (5, y20)$

is also on the graph of the implicitly defined function $y = y (x)$ .

A more efficient process uses a Maple for-loop that results in a list of points from which Maple's plot command can fashion a graph. This loop is implemented as follows.

>	$P ≔ NULL &colon; for k from 0 to 20 do X ≔ 2 + 3 / 20 k &semi; Y ≔ fsolve (eval (q, x = X), y) &semi; P ≔ P, [X, Y] &semi; end do &colon; P$

The NULL command at the beginning of the calculations starts an empty sequence of points. As each new $y$ -value is computed, the point $(x_{k}, y_{k})$ is formed and appended to the list.

A plot in which the computed points are linearly connected is then found in Figure 11.2.6.

>	$p1 := plot ([P], color = red) &colon;$ $p2 := plot ([P], style = point, symbol = solidcircle, color = black) &colon;$ $display ([p1, p2], labels = [x, y])$

Figure 11.2.4 Graph of $y (x)$ and the discrete points in Table 11.2.2

Exercises - Chapter 11

The equations given in Exercises 11.1 - 11.5 define one or more functions $y = y (x)$ implicitly. In each case,

(a) Using an appropriate technology, obtain a graph of the implicitly defined functions.

(b) Express explicitly each function in the graph plotted in Part (a).

11.1. $9 x^{2} + 10 y^{2} + 10 x - 2 y - 6 = 0$

11.2. $2 x^{2} - 9 y^{2} + 10 x - 3 y + 3 = 0$ , with $x$ in the interval $[- 10, 10]$ .

11.3. $2 x^{2} - 4 x y + y^{2} + 2 x - 12 y + 5 = 0$ , with $x$ in the interval $[- 15, 15]$ .

11.4. $7 x^{2} + 5 x y + 11 y^{2} + 9 x - 3 y - 7 = 0$

11.5. $5 x^{2} + 10 x y + 11 y^{2} + 10 x + y - 50 = 0$

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