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प्रश्न
Prove that the curves xy = 4 and x2 + y2 = 8 touch each other.
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उत्तर
Given circles are xy = 4 .....(i)
And x2 + y2 = 8 .....(ii)
Differentiating equation (i) w.r.t., x
`x * "dy"/"dx" + y * 1` = 0
⇒ `"dy"/"dx" = - y/x`
⇒ m1 = `- y/x` .....(iii)
Where, m1 is the slope of the tangent to the curve.
Differentiating equation (ii) w.r.t. x
`2x + 2y * "dy"/"dx"` = 0
⇒ `"dy"/"dx" = - x/y`
⇒ m2 = `- x/y`
Where, m2 is the slope of the tangent to the circle.
To find the point of contact of the two circles
m1 = m2
⇒ `- y/x = - x/y`
⇒ x2 = y2
Putting the value of y2 in equation (ii)
x2 + x2 = 8
⇒ 2x2 = 8
⇒ x2 = 4
∴ x = ± 2
∵ x2 = y2
⇒ y = ± 2
∴ The point of contact of the two circles are (2, 2) and (– 2, 2).
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