CONIC SECTIONS Introduction Sections of a cone Circle Parabola Ellipse Hyperbola

INTRODUCTION:Circles, ellipses, parabolas and hyperbolas are known as Conic Sections because they can be obtained as intersections of a plane with a double napped right circular cone.DOUBLE NAPPED RIGHT CIRCULAR CONELet l be a fixed vertical line & m be another line intersecting it at a fixed point V & inclined to it at an angle .Suppose we rotate the line m around the line l in such a way that the angle remains constant. Then the surface generated is a double napped right circular hollow cone herein after referred as double napped right circular cone.lm

The point V is called Vertex; the line l is axis of cone. The rotating line m is called a generator of cone. The vertex separates the cone into two parts is called nappes.Vml CONIC SECTION FROM A NAPPED RIGHT CIRCULAR CONE:If we take intersection of a plane with a cone, section so obtained is called conic section.

When = 90, the section is a circle.When < < 90, the section is a ellipse.When = , the section is a parabola.When 0 < , the plane cuts through both nappes & curves of intersection is a hyperbola.

When we throw a ball, the path covered by the ball is parabolic.This bridge is parabolic in nature.

Elliptical Orbits Around the Sun

A sand clock is hyperbolic in nature. The tyre of a car is circular in shape.

A circle is the set of all points in a plane that are equidistant from a fixed point in the plane..OP3P2P1centreRadius (r) Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point onthe circle .Then, by the definition, | CP | = r . By the distance formula,we have(x h) + (y k) = ri.e. (x h) + (y k) = rThis is the required equation of the circle with centre at (h,k) and radius r .XYO.C (h,k)P (x,y)OP1=OP2=OP3The fixed point is called the centre of the circle and the distance from the centreto a point on the circle is called the radius of the circle.

Q. Find the equation of the circle with centre (3, 2) and radius 4.Solution: Here h = 3, k = 2 and r = 4. Therefore, the equation of the required circle is (x + 3)+ (y 2) = 16Q. Find the centre and the radius of the circle x + y + 8x + 10y 8 = 0.Solution: The given equation is(x + 8x) + (y + 10y) = 8Now, completing the squares within the parenthesis, we get(x + 8x + 16) + (y + 10y + 25) = 8 + 16 + 25i.e. (x + 4) + (y + 5) = 49i.e. {x ( 4)} + {y (5)} = 72Therefore, the given circle has centre at ( 4, 5) and radius 7.Q. Find equation of circle passing through points (4,1) & (6,5) & whose centre is on the line 4x + y = 16.Solution: (x h) + (y k) = r(4 h) + (1 k) = r ------ (1) (6 h) + (5 k) = r ------ (2) 4h + k = 16 ------ (3)+ (2) (3), we get: -4h 8k = -44 4h + k = 16 -7k = -28 k = 4 , h= 3.

(4 3) + (1- 4) = r(1) + (-3) = rr = 10Equation of circle: (x 3) + (y 4) = 10Q. Does point (-4,2) lie inside, outside or on the circle x + y = 25 ?Solution: Distance between (-4,2) & (0,0)= ( -4 - 0) + (2 0) = 16 + 4 = 20 < 25 Thus, (-4,2) lies inside the circle.(4 3) + (1- 4) = r(1) + (-3) = rr = 10Equation of circle: (x 3) + (y 4) = 10Q. Does point (-4,2) lie inside, outside or on the circle x + y = 25 ?

Solution: Distance between (-4,2) & (0,0)= ( -4 - 0) + (2 0) = 16 + 4 = 20 < 25 Thus, (-4,2) lies inside the circle.

PARABOLAA parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane.The fixed line is called the directrix of the parabola. The fixed point F is called the focus. A line through the focus and perpendicular to the directrix is called the axis . The point of intersection of parabola with the axis is called the vertex.

Standard equations of parabola

The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. Four possible options for directrix, given vertex is at origin : ( x=a, x= -a, y = a, y = -a)

DERIVE PARABOLA FORMULALet F be the focus and l the directrix. Let FM be perpendicular to the directrix and bisect FM at the point O. Produce MO to X. Take O as origin, OX the x-axis and OY perpendicular to it as the y-axis. Let the distance from the directrix to the focus be 2a. Then, the coordinates of the focus are (a, 0), and the equation of the directrix is x + a = 0 as in Let P(x, y) be any point on the parabola such thatPF = PN, ... (1)where PN is perpendicular to l. The coordinates of B are ( a, y). By the distance formula, we havePF = (x a) + y and PB = (x + a) Since PF = PB, we have(x a) + y = (x + a) i.e. (x a) + y = (x + a) or x 2ax + a + y = x + 2ax + a or y = 4ax ( a > 0).

latus rectum (LL)vertex

1. Parabola is symmetric with respect to the axis of the parabola.If the equation has a y term, then the axis of symmetry is along the x-axis and if theequation has an x term, then the axis of symmetry is along the y-axis.2. When the axis of symmetry is along the x-axis the parabola opens to the(a) right if the coefficient of x is positive,(b) left if the coefficient of x is negative.3. When the axis of symmetry is along the y-axis the parabola opens(c) upwards if the coefficient of y is positive.(d) downwards if the coefficient of y is negative.

Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola. To find the Length of the latus rectum of the parabola y = 4axBy the definition of the parabola, AF = AC.But AC = FM = 2aHence AF = 2a.And since the parabola is symmetric with respect to x-axis AF = FB and soAB = Length of the latus rectum = 4a.

Q. Find the coordinates of the focus, axis,the equation of the directrix and latus rectum of the parabola y = 8x.Solution: Comparing with the given equationy = 4ax, we find that a = 2.Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is x = 2.Length of the latus rectum is 4a = 4 2 = 8.

Q. Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).Solution: Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form x = 4ay. Thus, we havex = 4(2)y, i.e., x = 8y.

ELLIPSEAn ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.The two fixed points are called the foci (plural of focus) of the ellipse.

The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse. We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b.

Take a point P at one end of the major axis.Sum of the distances of the point P to the foci is F1 P + F2P = F1O + OP + F2P(Since, F1P = F1O + OP) = c + a + a c = 2aTake a point Q at one end of the minor axis. Sum of the distances from the point Q to the foci is F1Q + F2Q = b + c + b+ c = 2 b + cSince both P and Q lies on the ellipse. By the definition of ellipse, we have2 b + c = 2a, i.e., a = b + cor a = b + c, i.e., c = a b.aF1F2RPQObcca- cb ab a

Eccentricity

The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by e) i.e., e = c/a.

Let PF1+PF2 = 2a where a > 0standard equation of an ellipse

Other form of Ellipselength of semi-major axis = alength of the semi-minor axis = blength of latus rectum = 2b/a.Latus rectumMinor axis = 2bMajor axis = 2a

A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.F1F2The two fixed points are called the foci of the hyperbola. The mid-point of the line segment joining the foci is called thecentre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola.Conjugate axisTransverse axisVerticesCentre

Let |PF1-PF2| = 2a where a > 0Standard equation of a hyperbola

length of latus rectum = 2b/alength of the semi-transverse axis = alength of the semi-conjugate axis = bTransverse axisLatus rectumConjugate axisEccentricity of hyperbola is e = c/aWhere a is the semi- transverse axis, b is the semi- conjugate axis & c is distance from the center to either focus.

OTHER FORM OF HYPERBOLA