Condition for Tangent to the Hyperbola Theorem Explained

In this overview, the condition for a line to touch a hyperbola is elucidated step by step. The equations of the line and hyperbola are analyzed to derive the necessary condition for the line to be a tangent to the hyperbola. The process involves finding the equation of the tangent at any point on the hyperbola, exploring point of contact, and more. The explanation is accompanied by mathematical derivations and solutions to aid in understanding the concept thoroughly.

• Hyperbola
• Tangent
• Theorem
• Co-ordinate Geometry
• Mathematics

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1. 2 D Co-ordinate Geometry Lecture-20 The hyperbola Dated:-19.05.2020 PPT-17 UG (B.Sc., Part-1) Dr. Md. Ataur Rahman Guest Faculty Department of Mathematics M.L. Arya, College, Kasba PURNEA UNIVERSITY, PURNIA

2. Condition for tangent to the hyperbola Theorem (1):-):-Find the condition that the Line should touch the hyperbola Sol:-The equations of the st.line and the hyperbola are given by = + = + y mx c 2 2 x a y b = 1. 2 2 Y=mx+c X ( , P x y ) 1 1 ....(1) c y b and y mx x a 2 2 + = 1....(2) and 2 2 (1) (2), From we get ( ) 2 + mx c 2 x a = 1 2 2 b + + = 2 2 2 2 2 2 2 2 ( 2 mca x ) b x b a m x a m mcx c a b + = 2 2 2 2 2 2 2 2 ( ) 2 ( ) 0...(3) x a c b

3. Continue If the line (1) touches the hyperbola (2), then the two values of x given by (3) must be equal. i.e the condition for which = 2 4 0 b ac ( ) = = + 2 + = 2 2 2 2 2 2 2 2 4( ) ( ) 0 mca b a m a c b 2 2 2 2 2 2 2 2 ( )( ) m c a b a m c b + 2 2 2 2 2 b c = 2 2 c a 2 2 2 2 4 m c a m a m b b 2 2 2 2 b a m c = + 2 2 2 ta n c a m b which is the required condition for ge ncy

4. Continue Remarks:-(i) Putting in eq. we get, is the eq. of the tangent to the hyperbola (ii) Point of Contact: Let contact of (1) and (2), then the eq. of tangent to the hyperbola (2) is 2 a Comparing x a m y and b = + = 2 2 2 y mx c c b a m = 2 2 2.....(4) 2 x a y mx b a m 2 y b = 1 . for all values of m 2 2 be the point of ( , P x y ) 1 1 xx yy b = 1....(5) 1 1 2 = (5) y b (1) . , with i e y x m mx c we get 2 1 c 1 c a m c = = = = 1 1 2 1 x 1 2 2 ( ) ( ) a b c b 2 1 c = ( = = = 2 2 2 , 1 2 y where c ) c c b a m 1 2 2 a m Hence Point of Contact is , c

5. Condition for tangent to the hyperbola Theorem (2):-Find the equation of the tangent at any point of the hyperbola Solution:- The eq. of the hyperbola is Let P and Q be two neighbouring points on the hyperbola (1). 2 2 1 1 2 2 1...(2) a b x y and a b Subtracting from we get x x y y a b y y x x b x x a y y + ( ) 2 2 x a y b , x y = 2 1 1 1 2 2 x a 2 y b = 1....(1) 2 2 Tangent at P ( , ) P x y x y 1 1 = 2 2 = 1....(3) 2 2 2 2 (2) (3), ( , ) Q x y 2 2 2 2 2 2 = 0 2 1 2 1 2 2 + 2 = .....(4) 2 1 2 1 2 2 1 2 1

6. Continue The eq. of the chord PQ passing through the points is 1 1 2 2 ( , ) ( , ) P x y and Q x y ( ) 1 1 2 1 2 2 1 1 1 2 2 1 ( ) ( ) x x x x y y y y a b y x y x = y y 2 1 x x + + x y x y b a + ( ) = . (4) y y x x from ( ) ( ) + = 1 2 1 1 2 1 0 2 2 When and PQ becomes the tangent at P Hence the eq. of the tangent at is ( ) 1 1 1 2 a xx yy x a b a Q P x x and y y 2 1 2 1 ( , P x y ) 1 ( 1 ) + + ( ) ( ) b x x x x y y y y = 1 1 1 0 2 2 2 y b xx a yy b = = + = 1 1 1 1 1 1 1 1 2 2 2 2 2 2