Equations and Solutions for Hyperbola Problems
Equations and solutions for hyperbola problems including finding equations, foci coordinates, latus rectum length, tangent equations, and more. Detailed step-by-step solutions provided.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.
E N D
Presentation Transcript
2 D Co-ordinate Geometry Lecture-21 The hyperbola Dated:-20.05.2020 PPT-19 UG (B.Sc., Part-1) Dr. Md. Ataur Rahman Guest Faculty Department of Mathematics M.L. Arya, College, Kasba PURNEA UNIVERSITY, PURNIA
Problems (1). Find the equation of the hyperbola whose focus (2,2) and directrix x+y=9 and its eccentricity is 2. (2) Find the axes, co-ordinates of the foci, length of latus- rectum, the eccentricity and the eq. of the directrix of the hyperbola (3) Show that the straight line touches the hyperbola Also find the point of contact. (4) Find the eq. of the tangent to the hyperbola which is parallel to the line (5) If are eccentricities of a hyperbola and its conjugate hyperbola, prove that e = 2 2 4 9 36. x y + = 21 5 116 x y = 2 2 7 5 232. x y e and e = 2 2 4 9 1 = + x y 4 5 7. y x 1 2 1 1 + = 1 2 2 e 1 2
Solution of (1) Solution:- Let S (2,2) be one focus, P(x, y) be any point on the hyperbola and PM be its distance from the directrix By the definition of hyperbola PS e PM PS PM PS PM = = + = = = 9...(1) 2 x y and eccentricity e = = 2 2 2 2 4 2 + 9 x y + + = 2 2 ( 2) ( 2) 4 x y 2 2 1 1 + + 81 2 + 2 2 4( 18 18 ) x y xy y x + + + = 2 2 ( 4 4) ( 4 4) x x y y 2 y + + = + + + 2 2 2 2 ( x 4 xy 4 32 8) 2 y 2 4 36 36 1 6 2 x y x y x + y = xy x + + 2 2 4 32 1 54 0 y x
Solution of (2) The Eq. of the given hyperbola is = 2 2 4 9 36 x y 2 2 4 36 x 9 x y = 1 36 2 2 y = 1....(1) 9 4 2 2 x a a y b and b = (1) 1...(2), Comparing with we get 2 2 = = = = 2 2 9 4 3 2, a and b Here a b + + 2 2 9 4 13 3 13,0) a b = = = = eccentricity e 3 a Co ordinates of foci are = ( ,0) ( ae
Solution continue Solution: Axes Length of transverse axis Length of conjugate axis Length of latus-rectum = 2 3 = = 2 6 a = 2 2 = = 2 4 b 2 2 2 4 3 8 3 b a = = = 3 13 3 9 13 a e Eq. of directrix is = = = x
Solution of (3) Solution:-The eqs. of the given st. line and the hyperbola are 21 5 21 116 1(116 21 7 (1) From and + = 116 x y = 5 x y = 5 )....(1) y x = 2 2 5 232,....(2) (2), we get and x y 2 1 21 = 2 7 (116 5 ) y 5 232 y 1 = 2 2 7 (116 5 ) y 5 232 y 21 21 + + + = 3 232 2 2 116 116 1160 4 ( 2) y 25 315 6 y y y + = = 2 4 0 0 y y = , 2 2 2 y
Solution Continue Since two values of y are equal. Therefore the line (1) touches the ellipse (2) 1 21 1 21 126 21 ( ) = (116 5 ) = 116 5 2 + = = (1), 6 From x y Hence the point of contact is ( ) 6, 2
Solution of (5) Solution:- Let the eq. of the hyperbola be 2 x a 2 a e a 2 y b b + = 1.....(1) where a b 2 2 + 2 2 1 a + Eccentricity Now, the conjugate hyperbola of is = = = ....(1) 1 2 2 2 e a b 1 2 2 2 2 x a y b y b x a = = 1 1.....(2) 2 2 2 2 + 2 2 a b = = eccentricity e 2 b 2 1 b + = ....(2) 2 2 2 e a b 2 (1) (2), Adding and a a we get 2 2 1 1 b + + = + = 1 + 2 2 2 2 2 2 e e b a b 1 1 2 1 + = 1 Pr ove d 2 2 e e 1 2
