Q. 2 C5.0( 1 Vote )

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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


4yx


...(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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