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TANGENTS AND NORMALS find the condition for the following set of curves to intersect orthogonally x^2/a^2 + y^2/b^2 = 1 and x^2/A^2 – y^2/B^2 =1

TANGENTS AND NORMALS
 
find the condition for the following set of curves to intersect orthogonally
x^2/a^2 + y^2/b^2 = 1 and x^2/A^2 – y^2/B^2 =1 

Grade:12

1 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
9 years ago
Given curves are
C1 : x^2/a^2 + y^2/b^2 = 1 .. and ... C2 : x^2/A^2 – y^2/B^2 =1
Suppose C1 and C2 intersect each other in the point P( x', y' ).
.........................................
Now, diff eq x^2/a^2 + y^2/b^2 = 1 of C1,
dy/dx = m1 ... so that ... slope of tgt to C1 at P(x', y') is

m1 = dy/dx.................. (1)
Next, differentiating eqx^2/A^2 – y^2/B^2 =1 of C2,

2x + 4y(dy/dx) = 0 ... so that ... slope of tgt to C2 at P(x',y') is

m2 = - (dy/dx) at P ......... (2)

From (1) and (2),

if m1·m2 = -1....... (3)

Then the curves are going to be orthogonal..

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