Analytic geometry. Textbook

The directrices are parallel to the y-axis. Since " > 1 we get that a" < a and the directrices do not intersect the hyperbola. The focus and the directrix, being on the same side from the y-axis, are called corresponding.

Theorem 10. The point belongs to the hyperbola if and only if the ratio of its focal distance to the distance of this point from the corresponding

directrix equals the eccentricity of the hyperbola.

Proof. We consider the left focus F1 = ( c; 0) and the left directrix de ned be the equation x = a" and denote by d1 the distance of a point M under consideration from this directrix.

1) Assume that M(x; y) is a point on the hyperbola. Then

are still going to prove that the point M is on the hyperbola, so we cannot use the formula for jF1Mj as in the previous case, but the formula for d1

p

(x + c)2 + y2 = j"x + aj:

Squaring the both sides of the last equation and swapping them we get:

("x + a)2 = (x + c)2 + y2;

"2x2 + 2"ax + a2 = x2 + 2cx + c2 + y2; ("2 1)x2 + 2("a c)x y2 = c2 a2:

We have proved that M is on the hyperbola.

The case of the right focus and directrix is analyzed similarly.

2.3.3Tangent to the hyperbola

Theorem 11. The tangent to the hyperbola xa2 yb2 = 1 at its point M0(x0; y0) may be de ned by the equation xa02x yb02y y = 1.

Proof. We consider the secant of the hyperbola passing through the

point M0 and a point M1 6= M0, M1(x1; y1). This secant has the equation

Multiplying the both sides of the equation (13) by the corresponding sides of the equality (14) and canceling we obtain:

canceling the equality by 2, opening the brackets and taking into account

Theorem 12. The straight line Ax + By + C = 0 is the tangent to the hyperbola xa22 yb22 = 1 if and only if A2a2 B2b2 = C2 and C 6= 0.

Proof. 1) Assume that the line Ax + By + C = 0 is the tangent to the given hyperbola at its point M0(x0; y0). Then the same line may be de ned by the equation xa20 x yb20 y = 1. Therefore

C2a2 C2b2 = 1; A2a2 B2b2 = C2:

2) Assume that there holds the equality A2a2 + B2b2 = C2 and C 6= 0. Let M0(x0; y0) be the point with the coordinates x0 = CA a2, y0 = BC b2 (the coordinates are taken from the rst part of the proof). Substituting these coordinates into the equation of the hyperbola and using the given relation we obtain that M0 is the point of the hyperbola. Using the previous theorem we get that the tangent to the hyperbola at this point may be de ned by the equation

or after simpli cation CAx + CBy = 1, Ax + By + C = 0. We obtain that the line de ned by the equation Ax + By + C = 0 is the tangent.

The case of the right focus and directrix is analyzed similarly.

Remark. There is valid the following statement. The straight line

Ax + By + C = 0 is the asymptote of the hyperbola xa22 yb22 = 1 if and only

These conditions are equivalent to the relations Aa = Bb (which are equivalent to A2a2 = B2b2) and C = 0.

Combining the statements of the previous theorem with this remark we obtain the following assertion.

Theorem 13. The straight line Ax + By + C = 0 is the tangent to the hyperbola xa2 yb2 = 1 or its asymptote if and only if A2a2 B2b2 = C2. If the last equality is valid then the given line is the tangent if and only if

C 6= 0 and the asymptote if and only if C = 0.

2.4The parabola

Definition. The parabola is the locus of the points on the plane each of which is equidistant from some xed point F of this plane and from some straight line d which does not pass through the point F . The point F is called the focus of this parabola, and the line d is called its directrix.

The obtained equation of the parabola is called its canonical (or standard) equation1). It is assumed that the eccentricity of parabola " = 1. From the equation (16) we get that for any point M(x; y) of the parabola there is valid the inequality x > 0. Therefore the parabola is located in the right half-plane.

If a point M(x0; y0) belongs to the parabola, then the point M1(x0; y0) also belongs to this parabola. It means that the graph of this curve is symmetric relative the x-axis, which is called the axis of the parabola. The intersection point of the parabola with its axis is called the vertex of the

parabola. In the rst quadrant, the equation of the parabola takes the form p

y = 2px . Therefore the value y monotonically increases as x ! +1 and tends to +1.

Theorem 14. The tangent to the parabola y2 = 2px at its point M0(x0; y0) may be de ned by the equation y0y = p(x + x0).

Proof. We consider the secant of the parabola passing through the given point M0 and a point M1 6= M0, M1(x1; y1). This secant has the equation

Since M0 and M1 are the points on the parabola, we have: y12 = 2px1; y02 = 2px0:

1)Since we squared nonnegative values there is no need to prove equivalence of the transformations.

Multiplying the both sides of the equation (17) by the corresponding sides of the equality (18) we obtain:

Now we assume that M1 ! M0. Then y1 ! y0 and passing to the limit in (19) we get: 2p(x x0) = 2y0(y y0). Canceling the equality by 2, removing the brackets, taking into account that y02 = 2px0 and swapping the sides of the equation we get:

px px0 = y0y y02; px px0 = y0y 2px0; y0y = px px0 + 2px0; y0y = p(x + x0):

Theorem 15. The straight line Ax + By + C = 0 is the tangent to the parabola y2 = 2px if and only if pB2 = 2AC.

Proof. 1) Assume that the line Ax + By + C = 0 is the tangent to the given parabola at its point M0(x0; y0). Then the same line can be de ned by the equation yy0 = p(x + x0) or px yy0 + px0 = 0. Therefore

Ap = By0 = pxC0 :

From here we get that A 6= 0 and x0 = CA , y0 = pBA . Since M0 is the point of the parabola we get:

2) Now we assume that there holds the equality pB2 = 2AC. Then

A 6= 0 since otherwise we get that B = 0, which is impossible since A

and B are the coe cients in the equation of the straight line. The point

M0(x0; y0) with the coordinates x0 = CA , y0 = pBA belongs to the parabola:

The tangent to the parabola at this point has the form

After simple transformations this equation takes the form Ax+By +C = 0. Therefore this line is the tangent.

2.5Polar equations of second order curves

Let us take one of the second order curves: ellipse which is not the circle, parabola or one branch of hyperbola. We choose a directrix of this curve. In the case of ellipse, we take any of its two directrices, in the case of hyperbola, we take the nearest directrix to the chosen branch. Let d be the chosen directrix and let F be the corresponding focus. We introduce the polar coordinate system on the plane such that F is the pole, the polar axis is orthogonal to the directrix d and is directed away from d. We denote by " the eccentricity of the curve under consideration.

Let A be the intersection point of the line containing the polar axis with the directrix. In the following we use the property which is valid for all second order curves listed above: the point is on the curve if and only if the ratio of its focal distance to the distance of this point from the corresponding directrix equals the eccentricity of the curve.

M

N d

A'

FK

Fig. 13. Derivation of the polar equation of the second order curve

Let us take the point N with the polar angle 2 . Let p be its polar radius. The distance of this point from the directrix d equals jAF j and we obtain

We take an arbitrary point M of the curve with the polar coordinates (r; ') and drop the perpendicular MK on the polar axis. The distance of M from the directrix equals jAKj and we obtain that

We get the desired polar equation

p

r = 1 " cos ': of the curve under consideration.

Remark. In the case of a circle (" = 0), the obtained equation (but not its derivation) remains correct.