Conic sections
1.circle:is the locus of a point which moves in a plane such that its
distance from a fixed point remains constant .the fixed point is called
the center and the constant distance is called the radius.
(x-h)2+(y-k)2=r2
x2+y2=r2
(standard form of the equation of circle)
(the circle passes through the origin)
* (general form of the equation of a conic section)
Ax2+By2+Dx+Ey+F=0 ;A=B=1; h=
(x-h)2+y2=r2 (center lies on x-axis)
x2+(y-k)2=r2 (center lies on y-axis)
(x-h)2+(y r)2=r2 (when circle touches x-axis)
(x- r)2+(y-k)2=r2 (when circle touches y-axis)
(x- r) 2+(y r)2=r2 (when circle touches both the co-ordinate axes)
concentric circles: two circles having the same center (h,k) but different
radii r1 and r2 are called concentric circles. Thus the circles :[(x-h)2+(y-k)2= r12]
[(x-h)2+(y-k)2=
………(2 )
Therefore the equations of concentric circles differ only in constant
terms.
Ex: find the center and radius of each of the following circles:
1.x2+(y-1)2=2
h=0 ,k=1 ,r=√
x2+(y-1)2= √
,center(0,1)
by comparing with (x-h)2+(y-k)2=r2
2. 4x2+4y2-10x+5y+5=0 (dividing by 4)
x2+y2 -
Comparing with Ax2+By2+Dx+Ey+F=0 ;h=
; k=
; r2=
√
Ex: find the equation of the circle which passes through the center of
the circle x2+y2+8x+10y-7=0 and is concentric with the circle
2x2+2y2-8x-12y-9=0?
x2+y2+8x+10y-7=0
By Comparing with Ax2+By2+Dx+Ey+F=0
h=
the center is (-4,-5)
equation of any circle concentric with the given circle
2x2+2y2-8x-12y-9=0 is 2x2+2y2-8x-12y+c=0 …………..(1)
Because it passes through (-4,-5) therefore:
2(16)+2(25)-8(-4)-12(-5)+c=0
2x2+2y2-8x-12y-174=0
c=-174 putting the value in (1)
or x2+y2-4x-6y-87=0
Ex: find the equation of circle whose center is (2,-1) and which passes
through the point (3,6)? Radius of circle =√
(x-h)2+(y-k)2=r2
(x-2)2+(y+1)2= √
x2+y2-4x+2y-45=0
√
Ex: find the equation of the circle whose diameters are 2x-3y+12=0 and
x+4y-5=0 and area=154?
The center of circle is the point of intersection of the diameters. solving
two equations we get the center of circle
[
]
The equation of circle
Ex: find the equation of circle tangent to the line y=4 and its center is
(2,7)?
y-4=0
Ax+By+C=0 ,A=0 B=1 C=-4
,if this line touches circle to find
the distance which equal to radius by using D=r=
√
/=/
(X-2)2+(Y-7)2=9
Ex:find the equation of circle passes through the points(4,2) ,(-6,2) and
its center lies on y-axis?
h=- =0 D=0
(4,2)
16+4+0+2E+F=0……(1)
(-6,-2)
36+4+0-2E+F=0 ……(2)
F=-30,E=5
X2+Y2+5Y-30=0
equation of circle
2.parabola:-A parabola is the locus of a point which moves
in a plane so that its distance from a fixed point in a plane
is equal to its distance from a fixed straight line in that
plane. The fixed point is called focus and the straight line
is the directrix of parabola.
√
X2=4PY
√
;e=eccentricity=1=c/a
Latus rectum of parabola:is the chord drawn through the focus
perpendicular to the axis of the parabola. Its length is 4p.
Ex findf the co-ordinates of the foci and the equation of directrix and
latus rectum length of
1. 3y2+16x=0
by comparing with y2=4px ;
Y2=
-
2. 2x2=5y
p=-
; focus (-4/3,0) , x=4/3 (directrix equation),length of
latus rectum=4p=16/3.
2. 2x2=5y
x2=
by comparing with x2=4py ;p=5/8
D y=-5/8 ; latus rectum length=4p=5/2
*Equation of parabola with no vertex the origin:
1.{focus on line parallel to y-axis} :(x-h)2=4p(y-k)
y=p+k
(x-h)2=-4p(y-k);D y=-p+k
2.Focus on line parallel to x-axis:(y-k)2=4p(x-h)
x=p+h
(y-k)2=-4p(x-h); D x=-p+h
Ex: find the focus ,vertex, directrix, of the following parabola:
1.y=-4x2+3x
2.4y2+12x-12y+39=0
1. y=-4x2+3x
y=-4(x2- x)
-1/4y=x2-3/4x
x2-3/4x+9/64=-1/4y+9/64
(x-3/8)2=-1/4(y-9/16) by comparing with (x-h)2=4p(y-k)
h=3/8,k=9/16,vertex(3/8,9/16); 4p=-1/4
p=-1/16
focus(3/8 ,9/16-1/16)=focus(3/8,1/2)
D y=1/16+9/16=5/8
2. 4y2+12x-12y+39=0 4(y2-3y)=- 12x-39;4(y2-3y
)=- 12x-39
4(y-3/2)2=-12x-39+9 4(y-3/2)2=-12x-30
(y-3/2)2=-3(x+5/2)
(
by comparing with(y-k)2=4p(x-h)
Vertex(-5/2,3/2)
4p=-3,p=-3/4;focus(-5/2-3/4,3/2)
D x=3/4-5/2=-7/4
) =-3x-15/2
focus(-13/4,3/2)