1 / 54

Mathematics

Mathematics. Session. Hyperbola Session - 1. Introduction. If S is the focus, ZZ´ is the directrix and P is any point on the hyperbola, then by definition. Question. Illustrative Problem.

sybil
Download Presentation

Mathematics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mathematics

  2. Session Hyperbola Session - 1

  3. Introduction If S is the focus, ZZ´ is the directrix and P is any point on the hyperbola, then by definition

  4. Question

  5. Illustrative Problem Find the equation of hyperbola whose focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity is 2. Solution : Let S(1, 2) be the focus and P(x, y) be any point on the hyperbola. where PM = perpendicular distance from P to directrix 3x + 4y + 8 = 0

  6. Solution Cont. Ans.

  7. Equation of The Hyperbola in Standard Form

  8. Definition of Special Points Lines of the Equation of Hyperbola (ii) Transverse and Conjugate Axes (i) Vertices (iii) Foci : As we have discussed earlier S(ae, 0) and S´(–ae, 0) are the foci of the hyperbola.

  9. (iv) Directrices :The lines zk and z´k´ are two directrices of the hyperbola and their equations are respectively. (vi) Eccentricity For the hyperbola we have Definition of Special Points Lines of the Equation of Hyperbola (v) Centre :The middle point O of AA´ bisects every chord of the hyperbola passing through it and is called the centre of the hyperbola.

  10. (ix) Focal Distance of a Point SP = ex – a S´P = ex + a Definition of Special Points Lines of the Equation of Hyperbola (vii) Ordinate and Double ordinate (viii) Latus rectum “A hyperbola is the locus of a point which moves in such a way that the difference of its distances from two fixed points (foci) is always constant.”

  11. The eccentricity of the conjugate hyperbola is Conjugate Hyperbola The conjugate hyperbola of the hyperbola

  12. Important Terms

  13. If is the equation Of hyperbola, then its auxiliary circle is x2 + y2 = a2 = eccentric angle are known as parametric equation of hyperbola. Auxiliary Circle and Eccentric AngleParametric Coordinate of Hyperbola The circle drawn on transverse axis of the hyperbola as diameter is called an auxiliary circle of the hyperbola.

  14. inside Outside inside Position of Point with respect to Hyperbola

  15. Let the equation of line is y = mx + c and equation of hyperbola is Intersection of a Line and a Hyperbola Point of intersection of line and hyperbola could be found out by solving the above two equations simultaneously.

  16. Intersection of a Line and a Hyperbola [Putting the value of y in the equation of Hyperbola] This is a quadratic equation in x and therefore gives two values of x which may be real and distinct, coincident or imaginary.

  17. Given hyperbola is and given line is y = mx + c Condition for Tangency and Equation of Tangent in Slope Form and Point of contact This is the required condition for tangency.

  18. Equation of Tangent in Slope Form Substituting the value of c in the equation y = mx + c, we get equation of tangent in slope form. Equation of tangent Point of Contact

  19. Equation of Tangent and Normal in Point Form Equation of tangent at any point (x1, y1)of the hyperbola is Equation of Normal at any point (x1, y1)of the hyperbola is

  20. Equation of tangent at is Equation of Tangent and Normal in Parametric Form Equation of normal in parametric form is

  21. Class Test

  22. Class Exercise - 1 Find the equation to the hyperbola for which eccentricity is 2, one of the focus is (2, 2) and corresponding directrix is x + y – 9 = 0.

  23. According to the definition of hyperbola Solution Let P(x, y) be any point of hyperbola. Let S(2, 2) be the focus. This is the required equation of hyperbola.

  24. Find the coordinates of centre, lengths of the axes, eccentricity, length of latus rectum, coordinates of foci, vertices and equation of directrices of the hyperbola Class Exercise - 2

  25. The given equation can be written as Solution

  26. The equation (i) becomes The coordinates of centre with respect to oldaxes are x – 1 = 0 and y – 2 = 0. Solution contd.. Shifting the origin at (1, 2) withoutrotating the coordinate axes, i.e. Put x – 1 = X and y – 2 = Y Centre: The coordinates of centre with respect to new axes are X = 0 and Y = 0.

  27. x = 1, y = 2 Solution contd.. Length of axes Length of transverse axes = 2b Length of conjugate axes = 2a Eccentricity

  28. Foci: Coordinates of foci with respect to new axes are, i.e.. Coordinates of foci with respect to old axes are(1, 5) and (1, –1). Vertices: The coordinates of vertices with respect tonew axes are X = 0 and , i.e. X = 0 and Solution contd.. Length of latus rectum

  29. The coordinates of axes with respect toold axes are x – 1 = 0, i.e. x = 1 and Vertices Directrices: The equation of directrices with respectto new axes are , i.e. . The equation of directrices with respect toold axesare , i.e. y = 3 and y = 1. Solution contd..

  30. Class Exercise - 3 • Find the equation of hyperbola whose • direction of axes are parallel to • coordinate axes if • vertices are (–8, –1) and (16, –1) and focus is (17, –1) and • focus is at (5, 12), vertex at (4, 2) and centre at (3, 2).

  31. (i)Centre of hyperbola is mid-point ofvertices Equation of hyperbola is Equation of hyperbola in new coordinate axes is. Solution Let x – 4 = X, y + 1 = Y.

  32. Solution contd.. As per definition of hyperbola a = Distance between centre and vertices = 144 Abscissae of focus in new coordinates system isX = ae, i.e. x – 4 = 12e

  33. Equation of hyperbola is (ii) Coordinates of centre are (3, 2). Equation of hyperbola is a = Distance between vertex and centre Equation (i) becomes Solution contd.. Let x – 3 = X, y – 2 = Y.

  34. Abscissae of focus is X = ae 5 = e + 3 [Abscissae of focus = 5] = 1 (4 – 1) = 3 Solution contd.. i.e. x – 3 = e (As a = 1) x = e + 3

  35. Class Exercise - 4 Find the equations of the tangents to the hyperbola 4x2 – 9y2 = 36 which are parallel to the line 5x – 3y = 2.

  36. Tangent is parallel to the given line 5x – 3y = 2 Equation of tangents Solution

  37. Find the locus of mid-point of portion of tangent intercepted between the axes for hyperbola Class Exercise - 5

  38. Any tangent to the hyperbolais be the middle point of AB Solution Let the tangent (i) intersect the x-axis at A and y-axis at B respectively. Let P(h, k) be the middle point of AB.

  39. Solution contd..

  40. Find the condition that the linelx + my + n = 0 will be normal to the hyperbola Class Exercise - 6

  41. Equations (i) and (ii) will represent the same line if Solution The equation of the given line is lx + my + n = 0 ...(ii)

  42. Solution contd..

  43. The curve represents • a hyperbola if k < 8 • an ellipse if k > 8 • a hyperbola if 8 < k < 12 • None of these Class Exercise - 7

  44. The given equation represents hyperbola if (12 – k) (8 – k) < 0 i.e. 8 < k < 12 Solution Hence, answer is (c).

  45. If the line touches the hyperbola at the point , show that Class Exercise - 8

  46. Solution

  47. Both (i) and (ii) represent same line Solution contd..

  48. Let where be two points on thehyperbola If (h, k) is the point of intersection of the normals at P and Q, then k is equal to (a) (b) (c) (d) Class Exercise - 9

  49. Solution

  50. From (iv) and (v) eliminating h, we get Solution contd..

More Related