Parabola

Coordinate Geometry

Parabola

1

Coordinate Geometry

Parabola ¤ Definition 1 (Conic Section). The locus of a point P , which moves so that its distance from a fixed point is always in a constant ratio to its perpendicular distance from a fixed straight line, is called a Conic Section. 1. The fixed point is called the Focus and is usually denoted by S. 2. The constant ratio is called the Eccentricity and is denoted by e. 3. The fixed line is called the Directrix. 4. The straight line passing through the focus and perpendicular to the directrix is called the Axis. Further, (a) When the eccentricity e is equal to unity, the conic section is called a Parabola. (b) When e is less than unity, the conic section is called an Ellipse. (c) When e is greater than unity, the conic section is called a Hyperbola. #I § 1. To find the equation of a Parabola. Let S be the fixed point and ZM the directrix (Fig. 1). We require the locus of a point P which moves in such a way that its distance from the point S be equal to P M , its perpendicular distance from ZM . Draw SZ perpendicular to the directrix and bisect SZ at the point A; produce ZA to x. The point A is clearly a point on the required curve. It is called the Vertex of the parabola. Take y Y P

M L

Z

A

N

S

X

x

L0 P0

Figure 1: Deriving the equation of a parabola. A as the origin, AX as the axis of x and AY , perpendicular to it, as the axis of y. Let the distance Anant Kumar

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ZA, or AS, be called a, and let P be any point on the curve whose coordinates are x and y. Join SP , and draw perpendicular respectively to the axis and the directrix. We have then SP 2 = P M 2 , i.e.

(x − a)2 + y 2 = ZN 2 = (a + x)2 , ∴

y 2 = 4ax

(1)

This being the relation which exists between the coordinates of any point P on the parabola and therefor is the required equation of the parabola. Equation 1 is the simplest possible equation and is also called the standard form of the equation describing a parabola. If instead of AX and AY we take the axis and the directrix ZM as the axes of coordinates, the equation would be (x − 2a)2 + y 2 = x2 ⇒ y 2 = 4a(x − a). Similarly, if the axis SY and a perpendicular line SL be taken as the axes of coordinates, the equation is x2 + y 2 = (x + 2a)2 ⇒ y 2 = 4a(x + a). Of course, the equation of the parabola referred to any focus and directrix can be written down by directly using the definition. ♣ Remark 1. The quantity y12 − 4ax1 is negative, zero, or positive accordingly as the point (x1 , y1 ) is within, upon, or without the parabola. ¤ Definition 2 (Latus Rectum). The latus rectum of any conic is the double ordinate drawn through the focus. In case of the parabola, the latus rectum is the segment LSL0 in Fig. 1. We have: SL = distance of L from the directrix = SZ = 2a. Hence, the length of the latus rectum = 4a. When the latus rectum is given, it follows that the equation of the parabola is completely known in its standard form, and the size and shape of the curve completely determined. For this reason, the quantity 4a is also often called the principal parameter of the curve. Focal Distance of any point P on the curve is its distance from the focus S. This distance SP = P M = ZN = ZA + AN = a + x. #I § 2. To find the points of intersection of any straight line with the parabola y 2 = 4ax. The given parabola: y 2 = 4ax.

(2)

y = mx + c.

(3)

and the equation of any straight line is

The coordinates of the points common to the straight line and the parabola satisfy both equations (2) and (3), and are therefore found by solving them. Substituting the expression for y from (3) in (2), we have (mx + c)2 = 4ax, ⇒ m2 x2 + 2x(mc − 2a) + c2 = 0

(4)

Eq. (4) is a quadratic equation for x and therefore has two roots, real, coincident, or imaginary. Anant Kumar

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The straight line therefore meets the parabola in two points, real, coincident, or imaginary. The roots of (4) are real or imaginary accordingly as 4(mc − 2a)2 − 4m2 c2 is positive or negative, i.e. accordingly as a(a − mc) is positive or negative, i.e. accordingly as mc ≶ a. #I § 3. To find the equation of the tangent at any point (x1 , y1 ) of the parabola y 2 = 4ax. We find the equation to the tangent at any point using the definition which was used in case of circle. Let P be the point (x1 , y1 ) and Q a point (x2 , y2 ) on the parabola. The equation to the secant P Q is y2 − y1 (x − x1 ). (5) y − y1 = x2 − x1 Since P and Q both lie on the parabola, we have y12 = 4ax1 , y22 = 4ax2 . Hence, by subtraction, we have y22 − y12 = 4a(x2 − x1 ), ⇒

(y2 − y1 )(y2 + y1 ) = 4a(x2 − x1 ), y2 − y1 4a ⇒ = . x2 − x1 y2 + y1

Substituting this expression in Equation 5, we have, as the equation to any secant P Q, 4a (x − x1 ) y2 + y1 y(y1 + y2 ) = 4ax + y1 y2 + y12 − 4ax1 y − y1 =

⇒

= 4ax + y1 y2

(6)

To obtain the equation of the tangent at (x1 , y1 ) we take Q indefinitely close to P , and finally put y2 = y1 . The Equation 6 then becomes 2yy1 = y12 + 4ax = 4ax + 4ax1 . Hence the equation of the tangent to the parabola y 2 = 4ax at the point (x1 , y1 ) lying on the parabola is: yy1 = 2a(x + x1 ) (7) #I § 4. To find the condition that the straight line y = mx + c may touch the parabola y 2 = 4ax. The line y = mx + c will touch the parabola y 2 = 4ax only if the roots of the quadratic given by Eq. (4) has repeated roots. This will happen provided 4(mc − 2a)2 = 4m2 c2 , ⇒

Anant Kumar

a2 − amc = 0, a so that c = . m

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Substituting this value in the equation of the straight line, we get that the straight line y = mx +

a

(8)

m

is always a tangent to the parabola y 2 = 4ax. In this equation, m is the tangent of the angle which the tangent makes with the axis of x. #I § 5. Equation to the normal to the parabola y 2 = 4ax at the point (x1 , y1 ) on the parabola. The required normal is the straight line which passes through the point (x1 , y1 ) and is perpendicular to the tangent at that point: 2a (x + x1 ). y= y1 The required equation is therefore y − y1 = m0 (x − x1 ), where m0 ×

2a = −1, y1

m0 = −

i.e.

y1 , 2a

and the equation of the normal is therefore y − y1 = −

y1 (x − x1 ). 2a

(9)

#I § 6. To express the equation of the normal in the form y = mx − 2am − am3 . In Eq. (9), put −

y1 = m, 2a

so that

Hence x1 =

y1 = −2am.

y12 = am2 . 4a

The normal is therefore y + 2am = m(x − am2 ), which is the same as y = mx − 2am − am3

(10)

and is normal at the point (am2 , −2am) of the curve. You should note that the “m” being talked of is the tangent of the angle which the normal makes with the axis. It must be carefully distinguished from the m of Equation 8 which the tangent makes with the axis. ¤ Definition 3 (Subtangent and Subnormal). If the tangents and normal at any point P of a conic section meet the axis in T and G respectively and P N be the ordinate at P , then N T is called the subtangent and N G the subnormal of P (Fig. 2). #I § 7. To find the length of the subtangent and the subnormal. If P be the point (x1 , y1 ), the equation of P T (Fig. 2) is, by Eq. (7), yy1 = 2a(x + x1 ).

(11)

To obtain the length of AT , we have to find the point where this straight line meets the axis of x, i.e. we put y = 0 in (11) and we have x = −x1 . (12) Anant Kumar

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y

P

A

T

N

G

x

Figure 2: The subtangent and the subnormal. Hence, AT = AN . The negative sign in Eq. (12) shows that T and N always lie on the opposite sides of the vertex A. Hence, the subtangent N T = 2AN = twice the abscissa of the point P . Since T GP is a right–angled triangle, we have P N 2 = T N · N G. Hence, the subnormal N G

PN2 PN2 = = 2a. TN 2AN Some Properties of the Parabola =

(13)

Refer to Fig. 3 for the properties. ¯ Property 1. If the tangent and normal at any point P of the parabola meet the axis in T and G respectively, then ST = SG = SP, and the tangent at P is equally inclined to the axis and the focal distance of P . Proof. Let P be the point (x0 , y 0 ). Draw P M perpendicular to the directrix. Then we have AT = AN. ∴ and hence Also,

T S = T A + AS = AN + ZA = ZN = M P = SP,

∠ST P = ∠SP T. N G = 2AS = ZS. ∴

SG = SN + N G = ZS + SN = M P = SP.

¯ Property 2. If the tangents at P meet the directrix in K, then KSP is a right angle. Proof. We have ∠SP T = ∠P T S = ∠KP M . Hence the two triangles KP S and KP M have the two sides KP , P S and the angle KP S equal respectively to the two sides KP , P M and the angle KP M . Hence, ∠KSP = ∠KM P = a right angle. Also ∠SKP = ∠M KP . Anant Kumar

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y M

P Y

K

T

A

Z M0

S

G

N

x

P0

Figure 3: Regarding the properties of the parabola. ¯ Property 3. Tangents at the the extremities of any focal chord intersect at right angles in the directrix. Proof. For, if P S be produced to meet the curve in P 0 , then, since ∠P 0 SK is a right angle, the tangent at P 0 meets the directrix in K. Also, by Property 2, ∠M KP = ∠SKP , and similarly, ∠M 0 KP 0 = ∠SKP 0 . Hence, 1 1 ∠P KP 0 = ∠SKM + ∠SKM 0 = a right angle. 2 2 ¯ Property 4. If SY be perpendicular to the tangent at P , then Y lies on the tangent at the vertex and SY 2 = AS · SP . Proof. The equation to any tangent is y = mx +

a . m

The equation to the perpendicular to this line passing through the focus is y=−

1 (x − a). m

These two lines meet where mx +

a 1 1 a = − (x − a) = − + , m m m m

that is where x = 0. Hence, Y lies on the tangent at the vertex. Further, if P be the point (x0 , y 0 ), then the equation of P T is yy 0 = 2a(x + x0 ). Hence |2a(a + x0 )| SY = p 4a2 + y 0 2 4a2 (a + x0 )2 2 (∵ y 0 = 4ax0 ) SY 2 = 4a2 + 4ax0 = a(a + x0 ) = AS · SP

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#I § 8. Through any given point (x1 , y1 ) there pass, in general, two tangents to the parabola y 2 = 4ax. The equation to any tangent is

a . m If this passes through the fixed point (x1 , y1 ), we have y = mx +

a m m2 x1 − my1 + a = 0

y1 = mx1 + i.e.

For any given values of x1 and y1 , this equation is in general a quadratic in m and hence gives two values of m (real or complex). Corresponding to each value of m we will have a different tangent. The roots of this quadratic will be real and distinct if y12 − 4ax1 is positive, i.e. the point (x1 , y1 ) lies without the curve. They are equal, i.e. the two tangents coincide, if y12 − 4ax1 becomes zero, that is if the point (x1 , y1 ) lies upon the curve. The two roots are imaginary if y12 − 4ax1 is negative, i.e. the point (x1 , y1 ) lies within the curve. #I § 9. Equation to the chord of contact of tangents drawn from a point (x1 , y1 ). The chord of contact is the line joining the points of tangency of the two tangents drawn from any external point. The equation to the tangent at any point Q whose coordinates are (x0 , y 0 ) is yy 0 = 2a(x + x0 ). Also the tangent at the point R with coordinates (x00 , y 00 ), is yy 00 = 2a(x + x00 ). If these tangents meet at the point T , whose coordinates are (x1 , y1 ), we have y1 y 0 = 2a(x + x0 ) and

00

00

y1 y = 2a(x + x )

(14) (15)

The equation of QR is then yy1 = 2a(x + x1 )

(16)

For, since (14) is true, the point (x0 , y 0 ) lies on (16). Also, since (15) is true, the point (x00 , y 00 ) lies on (16). Hence, (16) must be the equation to the straight line joining (x0 , y 0 ) to the point (x00 , y 00 ), i.e. it must be the equation to QR, the chord of contact of tangents from the point (x1 , y1 ). ¤ Definition 4 (Pole and Polar). If through a point P (within or without a parabola) there be drawn any straight line to meet the circle in Q and R, the locus of the point of intersection of the tangents at Q and R is called the polar of P with respect to the parabola; also P is called the pole of the polar. #I § 10. To find the equation of the polar of the point (x1 , y1 ) with respect to the parabola y 2 = 4ax.

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R0

T0 Q0 (x1 , y1 ) P Q

T (h, k)

R

Figure 4: Polar of a point with respect to a parabola. Let Q and R be the points in which any chord drawn through the point P (x1 , y1 ), meets the parabola (see Fig. 4). Let the tangents at Q and R meet in the point whose coordinates are (h, k). Then, we require the locus of (h, k). Since QR is the chord of contact of tangents from (h, k) its equation is ky = 2a(x + h) Since this straight line passes through the point (x1 , y1 ), we have ky1 = 2a(x1 + h) Since this relation is true, it follows that the point (h, k) always lies on the straight line yy1 = 2a(x + x1 )

(17)

Hence, Eq. (17) is the equation to the polar of (x1 , y1 ). #I § 11. Equation of the pair of tangents that can be drawn to the parabola from the point (x1 , y1 ). Let (h, k) be any point on either of the tangents drawn from (x1 , y1 ). The equation to the line joining (x1 , y1 ) to (h, k) is k − y1 (x − x1 ), h − x1 k − y1 hy1 − kx1 y = x+ . h − x1 h − x1

y − y1 = i.e.

If this be a tangent it must be of the form y = mx + so that

k − y1 = m and h − x1

a , m

hy1 − kx1 a = . h − x1 m

Hence, by multiplication, µ a= i.e. Anant Kumar

k − y1 h − x1

¶µ

hy1 − kx1 h − x1

¶ ,

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The locus of the point (h, k) (i.e. the pair of tangents required) is therefore a(x − x1 )2 = (y − y1 )(xy1 − yx1 ).

(18)

It will be seen that this equation can be written in an alternate form as (y 2 − 4ax)(y12 − 4ax1 ) = [yy1 − 2a(x + x1 )]2

(19)

#I § 12. To prove that the middle points of a system of parallel chords of a parabola all lie on a straight line which is parallel to the axis. Since the chords are all parallel, they all make the same angle with the axis of x. Let the tangent of this angle be m. The equation to QR, any one of these chords (Fig. 5), is therefore y = mx + c, where c is different for the several chords, but m is the same. The straight line meets the parabola y 2 = 4ax in points whose ordinates are given by my 2 = 4a(y − c)

⇒ y2 −

4a 4ac y+ = 0. m m

Let the roots of this equation, i.e. the ordinates of Q and R, be y 0 and y 00 , and let the coordinates of V , the middle point of QR be (h, k). Then 2a y 0 + y 00 = . 2 m The coordinates of V therefore satisfy the equation k=

2a , m so that the locus of V is a straight line parallel to the axis of the curve. y=

y Y

Q

P

T

V

A

X

x

R

Figure 5: A system of parallel chords. 2a 2a The straight line y = meets the curve in a point P , whose ordinate is and whose abscissa m m a is therefore 2 . m The tangent at this point is a y = mx + , m and is therefore parallel to each of the given chords. Hence the locus of the middle points of a system of parallel chords of a parabola is a straight line which is parallel to the axis and meets the curve at a point the tangent at which is parallel to the given system. Anant Kumar

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#I § 13. To find the equation to the chord of the parabola which is bisected at any point (h, k). By the last article the required chord is parallel to the tangent at the point P where a line through (h, k) parallel to the axis µ 2 meets ¶ the curve. k Thus the coordinates of P are , k . The tangent at this point is 4a µ ¶ k2 ky = 2a x + , 4a whose slope is 2a/k. The equation of the required chord is the equation of the straight line passing through the point (h, k) and parallel to above tangent. Thus the required chord is k(y − k) = 2a(x − h)

(20)

¤ Definition 5 (Diameter). The locus of the middle points of a system of parallel chords of a parabola is called a diameter and the chords are called its double ordinates. Thus in Fig. 5, P V is a diameter and QR and all the parallel chords are ordinates to this diameter. Thus, any diameter of a parabola is parallel to the axis and the tangent at the point where it meets the curve is parallel to its ordinates. #I § 14. The tangents at the ends of any chord meet on the diameter which bisects the chord. Let the equation of QR (Fig. 5) be y = mx + c,

(21)

and let the tangents at Q and R meet at the point (x1 , y1 ). Then QR is the chord of contact of tangents drawn from T , and hence its equation is yy1 = 2a(x + x1 ). Comparing this with equation (21), we have 2a = m, y1

so that y1 =

2a , m

and therefore T lies on the straight line

2a . m But this straight line is the diameter P V which bisects the chord. y=

Parametric form of the Parabola #I § 15. It is often convenient to express the coordinates of any point on the curve in terms of one variable. It is quite obvious that the values x=

a , m2

y=

2a m

always satisfy the standard equation to the parabola. Hence for all values of m, the point µ ¶ a 2a , m2 m Anant Kumar

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lies on the curve y 2 = 4ax. This m is equal to the tangent of the angle which the tangent at the point makes with the axis. The equation to the tangent at this point is y = mx +

a , m

and the normal is found to be my + x = 2a +

a . m2

#I § 16. The coordinates of the point could also be expressed in terms of the m of the normal at the point; in this case its coordinates are (am2 , −2am). The equation to the tangent at the point (am2 , −2am) is my + x + am2 = 0, and the equation to the normal is y = mx − 2am − am3 . #I § 17. The simplest substitution is x = at2 ,

and

y = 2at.

These values satisfy the equation y 2 = 4ax for all real values of t and they are taken as the parametric form of the standard parabola. The equation to the tangent and the normal at the point (at2 , 2at) are ty = x + at2 , y + tx = 2at + at3 .

and The equation to the straight line joining

(at21 , 2at1 ) and

(at22 , 2at2 )

is easily found to be y(t1 + t2 ) = 2x + 2at1 t2 . The tangents at the points (at21 , 2at1 ) and (at22 , 2at2 ) are t1 y = x + at21 , and

t2 y = x + at22 .

The point of intersection of these two tangents is clearly [at1 t2 , a(t1 + t2 )] . The points whose coordinates are (at2 , 2at) may be, for brevity, called the “point t”.

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#I § 18. To prove that, in general, three normals can be drawn from any point to the parabola and that the algebraic sum of the ordinates of the feet of these three normals is zero. The straight line y = mx − 2am − am3 is a normal to the parabola at the points whose coordinates are (am2 , −2am). If this normal passes through a fixed point O(h, k), we have y Y

P1 P2

A

X

x

O (h, k) P3

Figure 6: From a given point there can be drawn, in general, three normals. k = mh − 2am − am3 , i.e.

am3 + (2a − h)m + k = 0

(22)

This equation is a cubic equation in m and hence has three roots, real or imaginary. Corresponding to each of the three values of m, we will get each of the normals which passes through the fixed point O. Hence, three normals, real or imaginary, can be drawn through any point O. If m1 , m2 and m3 be the roots of (22), the sum of the roots m1 + m2 + m3 = 0. If the ordinates of the feet of these normals be y1 , y2 and y3 , we then have, y1 + y2 + y3 = −2a(m1 + m2 + m3 ) = 0. Hence, the second part of the proposition is proved. It turns out that for certain positions of the point O, all three normals are real; for other positions of O, one normal only will be real and the other two imaginary. There exists no other possibilities.

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