Q. 2 C5.0( 1 Vote )

# Show that the fol

Answer :

Given:

Curves x2 + 4y2 = 8 ...(1)

& x2 – 2y2 = 4 ...(2)

Solving (1) & (2),we get,

from 2nd curve,

x2 = 4 + 2y2

Substituting on x2 + 4y2 = 8,

4 + 2y2 + 4y2 = 8

6y2 = 4

y2 y = ± Substituting on y = ± , we get,

x2 = 4 + 2(± )2

x2 = 4 + 2( )

x2 = 4 + x2 x = ± x = ± The point of intersection of two curves ( , ) & ( , )

Now ,Differentiating curves (1) & (2) w.r.t x, we get

x2 + 4y2 = 8

2x + 8y. = 0

8y. = – 2x ...(3)

x2 – 2y2 = 4

2x – 4y. 0

x – 2y. 0

4y x ...(4)

At ( , ) in equation(3),we get   m1 At ( , ) in equation(4),we get     m2 = 1

when m1 & m2   × = – 1

Two curves x2 + 4y2 = 8 & x2 – 2y2 = 4 intersect orthogonally.

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