Parabolas in Coordinate Geometry

This lecture covers the basics of parabolas in coordinate geometry, including equations, vertices, focuses, directrices, latus rectum, and solving problems related to parabolas. Explore the properties and equations of parabolas to deepen your understanding of this fundamental concept in mathematics.

• Parabolas
• Coordinate Geometry
• Equations
• Vertices
• Focus
• Directrix

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1. 2 D Co-ordinate Geometry Lecture-10 Parabola UG (B.Sc., Part-1) By Dr. Md. Ataur Rahman Guest Faculty Department of Mathematics M.L. Arya, College, Kasba PURNEA UNIVERSITY, PURNIA

2. Parabola Vertex Focus Eq. of directrix Eq. of Axis Eq. of latus rectum Length of latus rectum 4a = x a = 2 4 , 0 y ax a (0,0) (a,0) = y = x a 0 (0,0) (-a,0) X=a X=-a 4a = Y=0 2 4 , 0 y ax a = 2 4 , ay a 0 x (0,0) (0,a) Y=-a X=0 Y=a 4a = 2 4 , ay a 0 x (0,0) (0,-a) Y=a X=0 Y=-a 4a

3. Problems based on Parabola Type I 1. Find the equation of the parabola whose focus is (3, 0) and directrix 2. Find the equation of the parabola with vertex at (0,0) and focus at(0, 2). 3. Find the equation of the parabola with vertex at the origin, the axis along the x-axis and passing through the point P(2,3). 4. Find the equation of the parabola with vertex at the origin, passing through the point P(3,-4) and symmetric about the y-axis. x = 3.

4. Solution of (1) = Given focus and the eq. of directrix is Since the focus lies on the x-axis, so x-axis is the axis of the parabola and focus (3,0) lies on the right hand side of the x-axis. So, it is a right- handed parabola. Then the equation of the required parabola is y (3,0) . 3 x = 3 ie a = 2 4 a = = x 2 4 3 2 y y x 2 1 x

5. Solution of (3) It is given that the vertex of the parabola at the origin and its axis lies along the x-axis. So, the eq. of the required parabola is But it passes through the point (2,3), so it lies in the first quadrant. So, it is a right-handed parabola. Then the eq. of the required parabola is Now, P(2,3) lies on (1) Hence, the required eq. is = = 2 2 4 4 y ax or y ax = 2 4 ...(1) 9 8 y ax = = 2 3 4 2 a a 9 8 9 2 = = = 2 4 4 y ax x x 9 2 = 2 y x

6. Problems based on Parabola Type II Find the co-ordinates of the focus, foot of the directrix and the vertex , the equation of directrix, axis and the latus rectum and length of latus rectum of the following parabolas = = = = 2 1. 2. 3. 4. 8 y y x x x 2 12 y x 2 6 2 16 y

7. Solution of (1) The equation of the given Parabola is = 2 8 y Here a x = = 4 8 2 a So, it is a right-handed parabola. So, (i) focus (2,0) lies on the positive direction of x-axis.. (ii)Vertex (0,0), (iii) foot of the directrix (-2,0) (iv) Eq. of directrix is Since X-axis is the axis of the parabola. (v) So, the eq. of axis (vi) The eq. of latus rectum (vii) The Length of latus rectum 2 . . ie x = + = 2 0 x y = 0 x a ie x a = = . 0 = 4 2 = = 4 8 a