*DEFINITION :*

__CONIC SECTION__

*A conic section or conic is the locus of a point, which moves in a plane so that its distance from a fixed point is in a constant ratio to its distance from a fixed straight line, not passing through the fixed point.*

*∎ The fixed point is called the focus.*

*∎ The fixed straight line is called the directrix.*

*∎ The constant ratio is called the eccentricity and is denoted by e.*

*∎ When the eccentricity is unity; i.e., e = 1 , the conic is called a parabola; *

*when e < 1, the conic is called an ellipse .*

*and when e > 1, the conic is called a hyperbola.*

*∎ The straight line passing through the focus and perpendicular to the directrix is called the axis of the parabola.*

*∎ A point of intersection of a conic with its axis is called vertex.*

__Standard equation of a Parabola:__

__Standard equation of a Parabola:__*Let S be the focus, ZM the directrix and P the moving point. Draw SZ perpendicular from S on the directrix. Then SZ is the axis of the parabola. Now the middle point of SZ , say A , will lie on the locus of P ,*

*i.e., AS = AZ*

*Take A as the origin, the x-axis along AS, and the y-axis along the perpendicular to AS at A, as in the figure.*

*Let AS = a , so that ZA is also a .*

*Let (x, y) be the coordinates of the moving point P.*

*Then MP = ZN = ZA + AN = a + x .*

*But by definition MP = PS*

*⇒ MP ^{2} = PS^{2}*

*So that , (a + x) ^{2} = (x – a)^{2} + y^{2}*

*Hence , the equation of parabola is y ^{2} = 4ax*

__Latus Rectum:__

__Latus Rectum:__*The chord of a parabola through the focus and perpendicular to the axis is called the latus rectum.*

*In the figure LSL’ is the latus rectum.*

*Also LSL’ = 4a = double ordinate through the focus S.*

*Note:*

*∎ Any chord of the parabola y ^{2} = 4ax perpendicular to the axis of the parabola is called double ordinate.*

*∎ Two parabolas are said to be equal when their latus recta are equal.**Four common forms of a Parabola*

*Four common forms of a Parabola:*

*Form : y ^{2} = 4ax *

*∎ Vertex : (0, 0)*

*∎ Focus : (a , 0)*

*∎ Equation of the Directrix: x = −a*

*Form: y ^{2} = −4ax *

*∎ Vertex : (0 , 0)*

*∎ Focus : (-a , 0)*

*∎ Equation of the Directrix : x = a*

*Form : x ^{2} = 4ay*

*∎ Vertex : (0 , 0)*

*∎ Focus : (0 , a)*

*∎ Equation of the Directrix : y = -a*

*Form: x ^{2} = -4ay*

*∎ Vertex : (0 , 0)*

*∎ Focus : (0 , -a)*

*∎ Equation of the Directrix : y = a*

*Illustration : Find the vertex , axis , directrix, focus, latus rectum and the tangent at vertex for the parabola 9y ^{2} − 16x − 12y − 57 = 0*

*Solution: The given equation can be rewritten as*

*which is of the form Y ^{2} = 4AX.*

*Hence the vertex is (− 61/16 , 2/3)*

*The axis is y − 2/3 = 0*

*⇒ y = 2/3*

*The directrix is : X + A = 0*

*⇒ x + 61/16 + 4/9 = 0*

*⇒ x = − 613/144*

*The focus is X = A and Y = 0*

*⇒ x + 61/16 = 4/9 and y − 2/3 = 0*

*⇒ (− 485/144 , 2/3) is the focus*

*Length of the latus rectum = 4A = 16/9*

*The tangent at the vertex is X = 0*

*⇒ x = − 61/16*

*Illustration : The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the equation of the parabola and the points where it meets the axes.*

*Solution: Focus of the parabola is the mid-point of the latus rectum.*

*⇒ S is (7, 4).*

*Also axis of the parabola is perpendicular to the latus rectum and passes through the focus. Its equation is*

*y − 4 = 0(x − 7)/(5−3)*

*⇒ y = 4*

*Length of the latus rectum = (5 − 3) = 2*

*Hence the vertex of the parabola is at a distance 2/4 = 0.5 from the focus. We have two parabolas, one concave rightwards and the other concave leftwards.*

*The vertex of the first parabola is (6.5, 4)*

*and its equation is (y − 4) ^{2} = 2(x − 6.5)*

*and it meets the x-axis at (14.5 , 0).*

*The equation of the second parabola is*

*(y − 4) ^{2} = −2(x − 7.5).*

*It meets the x-axis at (−0.5, 0) and the y-axis at (0, 4 ± √15).*

*Exercise :*

*(i) Find the equation of parabola whose focus is (1, −1) and whose vertex is (2, 1). Also find its axis and latus rectum.*

*(ii) Find vertex, focus, directix and latus rectum of the parabola y ^{2} + 4x + 4y − 3 = 0.*

*(iii) Find the equation of the parabola whose axis is parallel to the y – axis and which passes through the points (0, 4), (1, 9) and (−2, 6) and determine its latus rectum.*