Loci; Parabolas
Set of All Points
The set of all points that satisfy specified conditions is called the locus (plural: loci) of the point under the conditions.
EXAMPLE 37.1 The locus of a point with positive coordinates is the first quadrant $ (x > 0, y > 0) $ .
EXAMPLE 37.2 The locus of points with distance 3 from the origin is the circle $ x^{2} + y^{2} = 9 $ with center at (0, 0) and radius 3.
Distance Formulas
Distance formulas are often used in finding loci.
- DISTANCE BETWEEN TWO POINTS formula (derived in Chapter 8): The distance between two points $ P_{1}(x_{1},y_{1}) $ and $ P_{2}(x_{2},y_{2}) $ is given by
\[ d(P_{1},P_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} \]
- DISTANCE FROM A POINT TO A LINE formula: The distance from a point $ P_{1}(x_{1},y_{1}) $ to a straight line $ Ax + By + C = 0 $ is given by
\[ d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}} \]
EXAMPLE 37.3 Find the locus of points $ P(x, y) $ equidistant from $ P_{1}(1,0) $ and $ P_{2}(3,0) $ .
Set $ d(P,P_{1})=d(P,P_{2}) $ . Then $ = $ . Simplifying yields:
\[ (x-1)^{2}+(y-0)^{2}=(x-3)^{2}+(y-0)^{2} \]
\[ \begin{aligned}x^{2}-2x+1+y^{2}&=x^{2}-6x+9+y^{2}\\4x&=8\\x&=2\end{aligned} \]
The locus is a vertical line that forms the perpendicular bisector of $ P_{1}P_{2} $
Parabola
A parabola is defined as the locus of points P equidistant from a given point and a given line not containing the point, that is, such that PF = PD, where F is the given point, called the focus, and PD is the distance to the given line l, called the directrix. A line through the focus perpendicular to the directrix is called the axis (or axis of symmetry) and the point on the axis halfway between the directrix and the focus is called the vertex.
A parabola with axis parallel to one of the coordinate axes is said to be in standard orientation. If, in addition, the vertex of the parabola is at the origin, the parabola is said to be in one of four standard positions: opening right, opening left, opening up, and opening down.
Graphs of Parabolas in Standard Position
| OPENING RIGHT | OPENING LEFT | OPENING UP | OPENING DOWN |
| Vertex: $ (0,0) $ Focus: $ F(p,0) $ Directrix: x = -p | Vertex: $ (0,0) $ Focus: $ (-p,0) $ Directrix: x = p | Vertex: $ (0,0) $ Focus: $ F(0,p) $ Directrix: y = -p | Vertex: $ (0,0) $ Focus: $ F(0,-p) $ Directrix: y = p |
| Equation: $ y^{2} = 4px $ | Equation: $ y^{2} = -4px $ | Equation: $ x^{2} = 4py $ | Equation: $ x^{2} = -4py $ |
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| Figure 37-1 | Figure 37-2 | Figure 37-3 | Figure 37-4 |
Parabolas in Standard Orientation
Replacing x by x - h has the effect of shifting the graph of an equation by $ |h| $ units, to the right if h is positive, to the left if h is negative. Similarly, replacing y by y - k has the effect of shifting the graph by $ |k| $ units, up if k is positive and down if k is negative. The equations and characteristics of parabolas in standard orientation, but not necessarily in standard position, are shown in the following table.
| OPENING RIGHT | OPENING LEFT | OPENING UP | OPENING DOWN |
| Equation: $ (y - k)^{2} = 4p(x - h) $ | Equation: $ (y - k)^{2} = -4p(x - h) $ | Equation: $ (x - h)^{2} = 4p(y - k) $ | Equation: $ (x - h)^{2} = -4p(y - k) $ |
| Vertex: $ (h, k) $ | Vertex: $ (h, k) $ | Vertex: $ (h, k) $ | Vertex: $ (h, k) $ |
| Focus: $ F(h + p, k) $ | Focus: $ F(h - p, k) $ | Focus: $ F(h, k + p) $ | Focus: $ F(h, k - p) $ |
| Directrix: $ x = h - p $ | Directrix: $ x = h + p $ | Directrix: $ y = k - p $ | Directrix: $ y = k + p $ |
SOLVED PROBLEMS
37.1. Find the locus of points $ P(x,y) $ such that the distance of P from point $ P_{1}(2,0) $ is twice the distance of P from the origin.
Set $ d(P_{1},P)=2d(O,P) $ . Then $ =2 $ . Simplifying yields:
\[ (x-2)^{2}+y^{2}=4(x^{2}+y^{2}) \]
\[ x^{2}-4x+4+y^{2}=4x^{2}+4y^{2} \]
\[ 0=3x^{2}+3y^{2}+4x-4 \]
The locus is a circle with center on the x-axis.
37.2. Derive the formula $ d = $ for the perpendicular distance d from a point $ P_{1}(x_{1}, y_{1}) $ to a straight line $ Ax + By + C = 0 $ .
Drop a perpendicular from $ P_{1} $ to the line at point L. Then $ d = | | P(x, y) $ be an arbitrary point on the given straight line (see Fig. 37-5).
\[ PP_{1}L,d=\left|\overrightarrow{PP_{1}}\right|=\left|\overrightarrow{PP_{1}}\right|\cos\theta=\left|\overrightarrow{PP_{1}}\right|\frac{\overrightarrow{PP_{1}}\cdot\overrightarrow{PP_{1}}}{\left|\overrightarrow{PP_{1}}\right|\left|\overrightarrow{PP_{1}}\right|}=\frac{\overrightarrow{PP_{1}}\cdot\overrightarrow{PP_{1}}}{\left|\overrightarrow{PP_{1}}\right|\left|\overrightarrow{PP_{1}}\right|}\cdot \]
Now $ =x_{1}-x,y_{1}-y$ . To find $ $ , note that the given line has slope $ - $ , hence, any perpendicular line has slope $ $ . Therefore, every vector perpendicular to the given line, including $ $ , can be written as $ a,B/A$ for some value of a. Therefore,
\[ d=\frac{\overrightarrow{PP_{1}}\cdot\overrightarrow{P_{1}L}}{\left|\overrightarrow{P_{1}L}\right|}=\frac{\left\langle x_{1}-x,y_{1}-y\right\rangle.a\left\langle1,B/A\right\rangle}{\left|a\langle1,B/A\rangle\right|}=\frac{a[(x_{1}-x)+(B/A)(y_{1}-y)]}{\sqrt{a^{2}(1+B^{2}/A^{2})}} \]
Since the signs of a and A are unspecified, and a distance must be a nonnegative quantity, take the absolute value of the right-hand side to ensure that d is not calculated negative. Then
\[ d=\left|\frac{a[(x_{1}-x)+(B/A)(y_{1}-y)]}{\sqrt{a^{2}(1+B^{2}/A^{2})}}\right|=\left|\frac{a[A(x_{1}-x)+B(y_{1}-y)]}{a\sqrt{A^{2}+B^{2}}}\right|=\frac{|Ax_{1}-Ax+By_{1}-By|}{\sqrt{A^{2}+B^{2}}} \]
Finally, since $ (x,y) $ is on the line $ Ax + By + C = 0 $ , it must satisfy the equation of the line, hence the quantity -Ax - By can be replaced by C, and
\[ d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}} \]
37.3. Find the distance from (a) the point (5, -3) to the line $ 3x + 7y - 6 = 0 $ ; (b) the point (5, 7) to the line x = -4.
- Use the formula $ d = $ with $ x_{1} = 5 $ and $ y_{1} = -3 $ :
\[ d=\frac{\left|3\cdot5+7(-3)-6\right|}{\sqrt{3^{2}+7^{2}}}=\frac{12}{\sqrt{58}} \]
- Rewrite the equation of the line in standard form $ 1x + 0y + 4 = 0 $ , then use the formula $ d = $ with $ x_{1} = 5 $ and $ y_{1} = 7 $ :
\[ d=\frac{\left|1\cdot5+0\cdot7+4\right|}{\sqrt{1^{2}+0^{2}}}=9 \]
37.4. Show that the equation of a parabola with focus $ F(p, 0) $ and directrix x = -p can be written as $ y^{2} = 4px $ .
The parabola is defined by the relation PF = PD. Let P be an arbitrary point $ (x, y) $ on the parabola. Then PF is found from the distance-between-two-points formula to be $ PD $ is found from the formula for the distance from a point to a line to be $ |x + p| $ . Hence:
\[ \begin{aligned}PF&=PD\\\sqrt{(x-p)^{2}+(y-0)^{2}}&=|x+p|\ $ x-p)^{2}+y^{2}&=(x+p)^{2}\\x^{2}-2px+p^{2}+y^{2}&=x^{2}+2px+p^{2}\\y^{2}&=4px\end{aligned} \]
37.5. Show that the equation of a parabola with focus $ F(0, -p) $ and directrix y = p can be written as $ x^{2} = -4py $ . The parabola is defined by the relation PF = PD. Let P be an arbitrary point $ (x, y) $ on the parabola. Then PF is found from the distance-between-two-points formula to be $ $ . PD is found from the formula for the distance from a point to a line to be $ |y - p| $ . Hence:
\[ \begin{aligned}PF&=PD\\\sqrt{(x-0)^{2}+(y+p)^{2}}&=|y-p|\\x^{2}+(y+p)^{2}&=(y-p)^{2}\\x^{2}+y^{2}+2py+p^{2}&=y^{2}-2py+p^{2}\\x^{2}&=-4py\end{aligned} \]
37.6. For the parabola $ y^{2}=12x $ , find the focus, directrix, vertex, and axis, and sketch a graph.
The equation of the parabola is in the form $ y^{2}=4px $ with 4p=12, thus p=3. Hence the parabola is in standard position, with vertex $ (0,0) $ , opening right, and has focus at $ (3,0) $ , directrix the line x=-3, and axis the x-axis, y=0. The graph is shown in Fig. 37-6.
37.7. Show that $ y^{2}-8x+2y+9=0 $ is the equation of a parabola. Find the focus, directrix, vertex, and axis, and sketch a graph.
Complete the square on y to obtain:
\[ \begin{aligned}y^{2}+2y&=8x-9\\y^{2}+2y+1&=8x-8\ $ y+1)^{2}&=8(x-1)\end{aligned} \]
Thus the equation is the equation of a parabola in the form $ (y - k)^{2} = 4p(x - h) $ with p = 2, h = 1, and k = -1. Hence the parabola is in standard orientation, with vertex $ (1, -1) $ , opening right, and thus has its focus at
$ (h + p, k) = (2 + 1, -1) = (3, -1) $ . Its directrix is the line x = h - p = 1 - 2 = -1, and its axis is the line y = -1. The graph is shown in Fig. 37-7.
37.8. Find the equation of a parabola in standard position with focus $ (5,0) $ and directrix x = -5.
Since the parabola is in standard position with focus on the positive x-axis, the focus is located at the point $ (p, 0) $ , hence, p = 5. The parabola is opening right, hence its equation must be of form $ y^{2} = 4px $ . Substituting p = 5 yields $ y^{2} = 20x $ .
37.9. Find the equation of a parabola in standard orientation with focus $ (3, 4) $ and directrix the y-axis.
The equation can be found by direct substitution in the definition of the parabola PF = PD. Alternatively, note that the vertex is the point half the distance from the focus to the directrix, that is, the point $ (, 4) $ . Since the focus is to the right of the directrix, the parabola opens to the right and has an equation of the form $ (y - k)^{2} = 4p(x - h) $ , with $ h = $ and k = 4. The distance from the vertex at $ (, 4) $ to the focus at $ (3, 4) $ is then also $ $ , and this is the value of p. Substituting yields
\[ \begin{aligned}(y-4)^{2}&=4\Big(\frac{3}{2}\Big)\Big(x-\frac{3}{2}\Big)\ $ y-4)^{2}&=6x-9\end{aligned} \]
37.10. For the parabola $ x^{2} = -2y $ , find the focus, directrix, vertex, and axis, and sketch a graph.
The equation of the parabola is in the form $ x^{2} = -4py $ with 4p = 2, thus $ p = $ . Hence the parabola is in standard position, with vertex $ (0, 0) $ , opening down, and has focus at $ (0, -) $ , directrix the line $ y = $ , and axis the y-axis, x = 0. The graph is shown in Fig. 37-8.
37.11. Show that $ x^{2} + 2x + 6y - 11 = 0 $ is the equation of a parabola. Find the focus, directrix, vertex, and axis and sketch a graph.
Complete the square on x to obtain:
\[ \begin{aligned}x^{2}+2x&=-6y+11\\x^{2}+2x+1&=-6y+12\ $ x+1)^{2}&=-6(y-2)\end{aligned} \]
Thus the equation is the equation of a parabola in the form $ (x - h)^{2} = -4p(y - k) $ with $ p = $ , h = -1, and k = 2. Hence the parabola is in standard orientation, with vertex $ (-1, 2) $ , opening down, and thus has its focus at $ (h, k - p) = (-1, 2 - ) = (-1, ) $ . The directrix is the line $ y = k + p = 2 + = $ , and its axis is the line x = -1. The graph is shown in Fig. 37-9.
37.12. Find the equation of a parabola in standard position, opening down, with focus $ (0, -4) $ and directrix the line y = 4. Since the parabola is in standard position with focus on the negative y-axis, the focus is located at the point $ (0, -p) $ , hence p = 4. The parabola is opening down, hence its equation must be of form $ x^{2} = -4py $ . Substituting p = 4 yields $ x^{2} = -16y $ .
37.13. Find the equation of a parabola in standard orientation with focus (3,4) and directrix the line y = 6.
The equation can be found by direct substitution in the definition of the parabola PF = PD. Alternatively, note that the vertex is the point half the distance from the focus to the directrix, that is, the point $ (3,5) $ . Since the focus is below the directrix, the parabola opens down and has an equation of the form $ (x - h)^{2} = -4p(y - k) $ , with h = 3 and k = 5. The distance from the vertex at $ (3,5) $ to the focus at $ (3,6) $ is then 1, and this is the value of p. Substituting yields
\[ \begin{aligned}&(x-3)^{2}=-4(1)(y-5)\\&(x-3)^{2}=-4y+20\\ \end{aligned} \]
SUPPLEMENTARY PROBLEMS
37.14. Find the locus of points $ P(x,y) $ such that the distance from P to the y-axis is 5.
Ans. x = 5 and x = -5, two straight lines parallel to the y-axis.
37.15. Find the locus of points $ P(x,y) $ such that P is equidistant from both axes.
Ans. y = x and y = -x, two straight lines through the origin.
37.16. Find the locus of points $ P(x,y) $ such that the distance of P from $ P_{1}(1,1) $ is one-half the distance of P from $ P_{2}(-2,-2) $ .
Ans. $ x^{2} + y^{2} - 4x - 4y = 0 $ , a circle passing through the origin.
37.17. Find the locus of points $ P(x,y) $ equidistant from $ (5,-1) $ and $ (3,-8) $ .
Ans. $ 4x + 14y + 47 = 0 $ , a straight line, the perpendicular bisector of the line segment joining the given points
37.18. Find the locus of points $ P(x,y) $ equidistant from $ (-5,3) $ and $ x - y + 8 = 0 $ .
Ans. $ x^{2} + y^{2} + 2xy + 4x + 4y + 4 = 0 $ , that is, $ (x + y + 2)^{2} = 0 $ , a straight line perpendicular to the given line at the given point.
37.19. Find the locus of points $ P(x,y) $ such that the product of their distances from $ (0,4) $ and $ (0,-4) $ is 16.
\[ Ans.\quad x^{4}+2x^{2}y^{2}+y^{4}+32x^{2}-32y^{2}=0 \]
37.20. Show that the equation of a parabola with focus $ F(-p,0) $ and directrix x = p can be written as $ y^{2} = -4px $ .
37.21. Show that the equation of a parabola with focus $ F(0,p) $ and directrix y = -p can be written as $ x^{2} = 4py $ .
37.22. Sketch graphs of the equations (a) $ y^{2} = -2x $ ; (b) $ x^{2} = 6y $ .
Ans. (a) Fig. 37-10; (b) Fig. 37-11
37.23. Find equations for parabolas in standard position (a) with focus at $ (0,7) $ and directrix the line y = -7; (b) with focus at $ (-,0) $ and directrix the line $ x = $ .
Ans. (a) $ x^{2} = 28y $ ; (b) $ y^{2} = -5x $
37.24. Find equations for parabolas in standard orientation (a) with focus at $ (-2,3) $ and directrix the y-axis; (b) with focus at $ (-2,3) $ and directrix the line y = 1.
Ans. (a) $ y^{2}-6y+4x+13=0 $ ; (b) $ x^{2}+4x-4y+12=0 $
37.25. Sketch graphs of the equations (a) $ y^{2}-2y-3x-2=0 $ ; (b) $ x^{2}+2x+2y-5=0 $ .
Ans. (a) Fig. 37-12; (b) Fig. 37-13
37.26. Use the definition of the parabola directly to find the equation of a parabola with focus $ F(2,2) $ and directrix the line $ x + y + 2 = 0 $ .
\[ x^{2}-2xy+y^{2}-12x-12y+12=0 \]