Q. 2 C5.0( 1 Vote )

# Show that the following set of curves intersect orthogonally :

x^{2} + 4y^{2} = 8 and x^{2} – 2y^{2} = 4.

Answer :

Given:

Curves x^{2} + 4y^{2} = 8 ...(1)

& x^{2} – 2y^{2} = 4 ...(2)

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

from 2nd curve,

x^{2} = 4 + 2y^{2}

Substituting on x^{2} + 4y^{2} = 8,

⇒ 4 + 2y^{2} + 4y^{2} = 8

⇒ 6y^{2} = 4

⇒ y^{2}

⇒ y = ±

Substituting on y = ±, we get,

⇒ x^{2} = 4 + 2(±)^{2}

⇒ x^{2} = 4 + 2()

⇒ x^{2} = 4 +

⇒ x^{2}

⇒ x = ±

⇒ x = ±

∴ The point of intersection of two curves (,) & (,)

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

⇒ x^{2} + 4y^{2} = 8

⇒ 2x + 8y. = 0

⇒ 8y. = – 2x

...(3)

⇒ x^{2} – 2y^{2} = 4

⇒ 2x – 4y.0

⇒ x – 2y.0

⇒ 4yx

⇒ ...(4)

At (,) in equation(3),we get

m_{1}

At (,) in equation(4),we get

m_{2} = 1

when m_{1} & m_{2}

⇒ × = – 1

∴ Two curves x^{2} + 4y^{2} = 8 & x^{2} – 2y^{2} = 4 intersect orthogonally.

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