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CONIC SECTIONS LOCI The curve described by a point which moves under given conditions is called its locus (plural loci). Example : If C (h,k) is a fixed point in the Cartesian plane and P (x,y) is a variable point (moving) on that plane such that its distance from C is always equal to its y coordinate, find the locus of P. Solution :

Therefore, distance PC = y,

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w c2 ` a2 b s [ x @ h + y @ k = y,

d

`

a

`

a

distance between two points P1 x1 , y1 and P 2 x 2 , y 2 iε A P1 P 2

[ x @h + y@k `

a2 b

c2

= y2

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w we ` a2 ` a2 q = x @x + y @ y 1

2

1

2

[ x 2 @ 2hx + h + y 2 @ 2ky + k = y 2 2

2

[ x 2 @ 2hx + h = 2ky @ k 2

2

kf f f [ x @ h = 2k y @ f which is the equation of a parabola A 2 `

a2

f

g

CONIC SECTIONS Conic sections are the curves that are formed when we make a straight cut in a cone i.e. when a plane intersects a right circular cone. These curves are • Circle, • Parabola • Ellipse and • Hyperbola.

CIRCLE If in a plane there is a fixed point C (h,k) and a point P (x,y) moves in such a way that its distance from C is always constant, then the path or locus of P is a circle. If the constant distance = r Then d (P,C) = r. Using distance formula w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w

b c2 s` x @ ha2 + y @ k =r

[ x @h + y@k `

a2 b

c2

= r2

The standard equation of the circle with centre (h,k) and radius r is given by `

x @h + y@k a2 b

c2

= r2

If the centre of the circle is the origin (0,0) and the radius is r, then the equation becomes x 2 + y 2 = r2

Example : Find the equation of the circle with centre (3,-2) and radius 5. Solution : Using h = 3, k = -2 in the standard equation we get

`

aC2

x @3 + y@ @2 a2 B

a2 b

`

c2

=5

[ x @3 + y + 2 =5 `

2

2

Example : Find the centre and radius of the circle x 2 @ 4x + y 2 + 2y @ 9 = 0 Solution :

x 2 @ 4x + y 2 + 2y @ 11 = 0 x 2 @ 4x + y 2 + 2y = 11 b `

x 2 @ 4x + 4 + y 2 + 2y + 1 = 11 + 4 + 1 c b

c2

c

x @2 + y + 1 =4 a2 b

2

Comparing with the standard equation Centre 2, @ 1 Radius = 4 units b

c

PARABOLA If in a plane there is a fixed point F and a fixed line l, and a point P moves in such a way that its distance from F and l is always equal, then the path or locus of P is a parabola. F is called the Focus and l is the Directrix. The line through focus perpendicular to the directrix is the axis of symmetry. The vertex V lies on this axis halfway between the focus and the directrix or the point at which the parabola intersects the axis of symmetry.

Out of all the possible parabolas, we restrict ourselves to those, which have vertex at the origin and the axis of symmetry is either vertical or horizontal (i.e. it is either the x-axis

or y-axis). In these, four standard positions are possible: a) opening up, b) opening down, c) opening right, d) opening left.

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4. Opening Right Equation : y 2 = 4px Vertex : (0,0) Focus : F (p,0) Directrix : x = -p Fig. 1.

Opening Left Equation : y 2 = @ 4px Vertex : (0,0) Focus : F (-p,0) Directrix : x = p Fig. 2.

Opening Up Equation : x 2 = 4py Vertex : (0,0) Focus : F (0,p) Directrix : y = -p Fig. 3.

Opening Down Equation : x 2 = @ 4py Vertex : (0,0) Focus : F (0,-p) Directrix : y = p Fig. 4.

Other forms of Parabolas : Equations of parabolas whose axis of symmetry is either vertical or horizontal, but the vertex is not the origin is discussed here. Let the vertex be V(h,k). We can get the equations by replacing x with x-h and k with y-k. Opening Right Equation : b

c2

Opening Left Equation :

y @ k = 4p x @ h `

Vertex : (h,k) Focus : F (h+p,k) Directrix : x =h -p

b

a

c2

y @ k = @ 4p x @ h `

Vertex : (h,k) Focus : F (h-p,k) Directrix : x =h+ p

Opening Up Equation : b c ` a2 x @ h = 4p y @ k

a

Opening Down Equation : b c ` a2 x @ h = @ 4p y @ k

Vertex : (h,k) Focus : F (h,k+p) Directrix : y =k -p

Vertex : (h,k) Focus : F (h,k-p) Directrix : y =k+ p

Example : Show that y 2 @ 8x + 2y + 9 = 0 is the equation of a parabola. Also find its focus, vertex, directrix. Solution :. y 2 @ 8x + 2y + 9 = 0 y 2 + 2y + 1 = 8x @ 8 c2

b

y + 1 = 4.2 x @ 1

b

y@ @1

`

ac2

`

a

= 4.2 x @ 1 `

a

b

comparing this equation with y @ k p = 2, h = 1, b

c

Vertex is 1, @ 1 ,

k =@1

b

c2 c

= 4p x @ h , we get

Focus is 3, @ 1 ,

`

a

Directrix is x = @ 1

ELLIPSE If in a plane there are two fixed points F1 and F2, and a point P moves in such a way that the sum of its distances from these points is constant, then the path or locus of P is an ellipse.

The points F1 and F2 are called foci (plural of focus). Then PF1+PF2 = 2a. (2a is a positive constant).

For the simplest ellipse, we take the foci on the x-axis at (c,0) and (-c,0) and the sum of the distances as 2a. We can make the equation of the ellipse by using the distance formula. After simplification the standard equation of such an ellipse is 2 2 xf f f f f f f yf f f f f f f + =1 2 2 a b

where a>b and b = a 2 @ c 2 2

In this case the foci are (c,0) and (-c,0). The mid point of these is the origin (0,0) which is called the centre. This ellipse intersects the x-axis at (a,0) and(-a,0). These points are called vertices. The segment that joins these two points is called the major axis. The length of the major axis is 2a. This ellipse intersects the y-axis at (0,b) and(0,-b). The segment that joins these two points is called the minor axis. The length of the minor axis is 2b. Equation

Foci

2 2 xf f f f f f f yf f f f f f f + =1 2 2 a b where b ca>b>0 2 F c,0 , c 2 = a 2 @ b ,

Foci on the x-axis

2 2 xf f f f f f f yf f f f f f f + =1 2 2 a b bwhere ca>b>0 2 0, F c , c 2 = a 2 @ b

Foci on the y-axis

Vertices Centre Major axis Minor axis

b

c

b

F a,0

c

0, F a

At Origin (0,0) Horizontal, length 2a Vertical, length 2b

At Origin (0,0) Vertical, length 2a Horizontal, length 2b

Other forms of Ellipses: We can make the equations of the ellipses whose axes are parallel to the x-axis and yaxis but whose centre is not at the origin(0,0). Let the centre be (h,k). We can get the equations by replacing x with x-h and k with y-k.

Equation

b c2 ` a2 y @ k xf @ hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

+ 2 a2 b where a>b>0

=1

Foci F c + h,k , c 2 = a 2 @ b

b

Vertices Centre Major axis Minor axis

c

, Foci on thecx-axis b F a + h,k

At (h,k) Horizontal, length 2a Vertical, length 2b

2

b c2 ` a2 y @ k xf @ hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

+ =1 2 a2 b bwhere a>b>0 c 2 h, F c + k , c 2 = a 2 @ b

Foci on the y-axis

b

h, F a + k

c

At (h,k) Vertical, length 2a Horizontal, length 2b

Have you noticed that when the centre shifted to (h,k) with the axes parallel to the coordinate axes, we have added h to the x-coordinates and k to the y-coordinates for the vertex, foci etc, keeping the lengths of the axes unchanged. Example : Find the foci, vertices, and the lengths of the major and minor axes of the ellipse given by the equation 2 2 xf f f f f f f yf f f f f f + f =1 25 9 Solution : As the denominator of x2 is greater, the ellipse has horizontal major axis. 2 2 xf f f f f f yf f f f f f f So, the equation is of the type f + = 1 where a>b>0 a2 b2 2 2 a 2 = 25, b = 9, c 2 = a 2 @ b = 25 @ 9 = 16 #a=5 b = 3, c = 4, b

Foci

c

F 4,0

b

Vertices

c

F 5,0

Length of major axis Length of minor axis

10 8

Example : Find the foci of the ellipse 9x 2 + 4 y 2 = 36 . Solution :

9x 2 + 4 y 2 = 36 2 2 xf f f f f f f yf f f f f f + f =1 4 9

dividing both sides by 36 we get

2 2 xf f f f f f yf f f f f f f + =1 this is of the type f 2 2 a b 2 # c2 = a2 @ b = 9 @ 4 = 5 w w w w w w [ c = p5

b

where b = 4, a 2 = 9 2

w w w w w wc

The foci are 0, F p5 A

HYPERBOLA If in a plane there are two fixed points F1 and F2, and a point P moves in such a way that the difference of its distances from these points is constant, then the path or locus of P is a hyperbola. The points F1 and F2 are called foci (plural of focus). Then PF1 - PF2 = F 2a (2a is a positive constant).

For the simplest hyperbola, we take the foci on x-axis at (c,0) and (-c,0) and the difference of the distances as 2a. We can make the equation of the hyperbola by using the distance formula. After simplification the standard equation of such an hyperbola is 2 2 xf f f f f f f yf f f f f f f @ =1 a2 b2

where a>0, b>0 and c 2 = a 2 + b

2

In this case the foci are (c,0) and (-c,0). The mid point of these is the origin (0,0) which is called the centre. This hyperbola intersects the x-axis at (a,0) and(-a,0). These points are called vertices. The segment that joins these two points is called the transverse axis. The length of the transverse axis is 2a. This hyperbola does not intersect the y-axis. Still we have shown the points (0,b) and (0,b). The line segment joining these two points is called conjugate axis. Equation

Foci

2 2 xf f f f f f f yf f f f f f f @ =1 2 2 a b where a>0,b>0 b c 2 F c,0 , c 2 = a 2 + b ,

Foci on the x-axis

2 2 xf f f f f f f yf f f f f f f @ =1 2 2 a b where ca>0, b>0 b 2 0, F c , c 2 = a 2 + b

Foci on the y-axis

b

Vertices Centre Asymptotes Transverse axis Conjugate axis

c

F a,0

At Origin (0,0) bf f f y =F f x a Horizontal, length 2a Vertical, length 2b

b

c

0, F a

At Origin (0,0) af f f y =F f x b Vertical, length 2a Horizontal, length 2b

Other forms of Hyperbolas: We can make the equations of the hyperbolas whose axes are parallel to the x-axis and yaxis but whose centre is not at the origin(0,0). Let the centre be (h,k). We can get the equations by replacing x with x-h and k with y-k. Equation

Foci

Vertices Centre Asymptotes Transverse axis Conjugate axis

b c2 ` a2 y @ b x @ a f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

@ =1 2 a2 b where a>0,b>0 b c 2 F c + h,k , c 2 = a 2 + b

, Foci on thecx-axis b F a + h,k

At (h,k)

` a bf f f y @k =F f x @h a Horizontal, length 2a Vertical, length 2b

b c2 ` a2 y @ b x @ a f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

@ =1 2 a2 b where a>0,c b>0 b 2 h, F c + k , c 2 = a 2 + b

Foci on the y-axis b

h, F a + k

c

At (h,k) ` a af f f y @k =F f x @h b Vertical, length 2a Horizontal, length 2b

When the centre shifted to (h,k) with the axes parallel to the coordinate axes, we have added h to the x-coordinates and k to the y-coordinates for the vertex, foci etc, keeping the lengths of the axes unchanged. Example : Find the foci, vertices, and the asymptotes of the hyperbola given by the equation 9x 2 @ 25 y 2 = 225 Solution :

9x 2 @ 25 y 2 = 225 dividing both sides by 225 we get 2 2 xf f f f f f f yf f f f f f @ f =1 25 9 As the x 2 term is positive , the hyperbola has horizontal transverse axis A a 2 = 25, #a=5

b = 9, b = 3, 2

Foci Vertices Equation of the asymptotes

c 2 = a 2 + b = 25 + 9 = 34 w w w w w w w w w w w c = p34 , 2

b

w w w w w w w w w w w c

F p34 ,0

b

c

F 5,0

3f 3f f f y= f x and y = @ f x 5 5