The point (2,-1) lies on the required tangent, so that if this is to be its equation we must have

-1-4 m (2+2)+5V1+m2 or -5-4m-±5V1+m2.

Squaring and collecting terms,

9m2-40m=0;

so that for the required tangent m=0 or 40. The two tangents are thus

and

y=-1

y+1=40(x-2) or 40x-9y-89 0.

Examples.

1. Find the tangents to x2+y2+8x-10y+1=0 having the slope; trace the circle and the tangents.

2. Find the tangents to the circle of Example 1 from the point (3, -4); trace the circle and the tangents. 3. Find the tangents to x2 + y2 6х (-1,-2); trace the circle and the tangents. 4. Trace in one figure the lines represented by

10y 180 from

y-2=m(x-1)±2√1+m2

5

when m= −y, −1, −1, −2, 0, 12, 4, 4, 12, respectively, and draw the circle that is tangent to all of them.

119. Loci of Quadratic Equations.-The most general form of quadratic equation in two variables is

ax2+bxy+cy2+dx+ey+f=0.

(1)

If some particular value x1 is given to x, such an equation becomes a quadratic in one unknown quantity, y, which can be solved. This is equivalent to finding the intersection of the locus of (1) with the line x=x,; generally two points in this vertical line are found, but they may be imaginary, indicating that the line does not cut the locus, or coincident, indicating that the line is tangent to the locus.

Enough points to plot the locus might be found in this way, but the process would be very tedious. It is much simpler to solve Equation (1) for y in terms of x once for all, and substitute various values of x in the result. In this process, certain principles become evident which shorten the work.

The procedure varies according as the factors of the terms of highest degree, ax2+bxy+cy2, are imaginary, real and equal, or real and different; that is, according to whether b2-4ac is less than zero, equal to zero, or greater than zero. The loci in these three cases are given the names of ellipse, parabola, and hyperbola, respectively.

120. The Ellipse.-Given the equation

2x2-2xy + y2+2x−4y+1=0,

for which b2-4ac-4, to find its locus.

(1)

First solve for y, thus expressing y as an explicit function of x:

for any value we may give to x, we get one point on this line, and two points on the graph of (2), one V(3-x) (x+1) vertically above the point on (3), the other √(3−x) (x+1) vertically below. The curve of (2) consequently has a vertical chord 2V (3-x)(x+1) in length, bisected by y=x+2. As this is true for every value of x, the line y=x+2 bisects every vertical chord of (2), and so is called a diameter of the ellipse.

It is evident that we shall get real values of y from (2) only when x is less than 3 and greater than -1; there is therefore no part of the curve to the right of the vertical line x=3, or to the left of the vertical line x=-1. Moreover, when x=3 or -1, the two values of y become coincident, and the length of the vertical chord becomes zero; i. e., x=3 and x=-1 are tangent to the curve. They are called the vertical limiting tangents.

The length of the vertical chord, 2√(3−x)(x+1), has its greatest value when the value of x is half-way between the limiting values, 3 and 1, or when x=1. The extremities of this -1, maximum vertical chord have the coordinates

x=1 and y=1+2±√(3−1)(1+1)=3±2=5 or 1; that is, are the points (1, 5) and (1, 1), the length of the chord being 4.

It can be shown that the vertical line x=1 is another diameter of the ellipse, bisecting all chords parallel to y=x+2; the two diameters are said to be conjugate. Their intersection is the center of the ellipse, the point (1,3).

It can be shown that parallels to y=x+2 (the conjugate to the vertical diameter) drawn through the ends of the vertical diameter are tangent to the ellipse.

We have thus located a parallelogram such that the ellipse is tangent to each of its sides at the middle point. This parallelogram determines the ellipse completely, and is sufficient for what is called tracing the ellipse; i. e., sketching a smooth oval having the properties so far described.

121. Tracing an Ellipse.-The steps of the process may be summarized:

Given any equation Ax2+Bxy+Cy2+Dx+Ey+F=0 having B2-4AC <0:

(1) Solve for y.

(2) Trace the oblique diameter.

(3) Draw the vertical limiting tangents.

(4) Draw the vertical diameter, locate its extremities, determine its length, and locate the center.

(5) Draw parallels to the oblique diameter through the extremities of the vertical diameter.

(6) Sketch the ellipse, tangent to the parallelogram at the ends of the two conjugate diameters.

It is possible to solve similarly for x in terms of y and get horizontal limiting tangents, the horizontal diameter and the diameter conjugate to it.

122. Construction of an Ellipse.-It can be proved that the following construction will give points of an ellipse.

Given a parallelogram ABCD, Fig. 22, divided into four parallelograms by lines FGF, K'GK drawn through its center, G, parallel to its sides. Divide the line FG into any number of equal parts, and number the points of division consecutively, beginning with F. Divide FC into the same number of equal parts and number the points of division, beginning at F.

Join K with any point of FC, and K' with the corresponding point of FG, the two lines intersecting at P; then P is a point of the ellipse tangent to the sides of the parallelogram at F", K, F and K'.

Treating the other corners of the parallelogram in the same way, we can find any number of points of the ellipse. The four points shown in the figure are especially useful and very easily found.

Examples.

1. Trace 4y-4xy+5x2-8y+12x-28-0, solving for y. 2. Trace 16y2+24xy+25x2-96y-136x+64-0, solving for y and x both.

3. For any point on the ellipse of Example 1,

What is the slope of the ellipse at its intersections with 2y-5x -6=0? With 2y-x-2=0? What are these lines? What is their intersection?

4. Find the locus of points as far from the point (2, 2) as from the line x+y=0, and trace it.

5. In the ellipses of Examples 1 and 2, find the slope of the tangents at the ends of the vertical diameter, showing it to be the same as the slope of the diameter conjugate to the vertical.

6. Given the ellipse 3x2-xy+y2+6x+6y=0; find its intersections with the axes, and find and draw the tangent at each. Draw the line which will pass through the highest and lowest points of the ellipse, and the line which will pass through the extreme right and left points. Sketch the ellipse.