Prove that the curves xy = 4 and $x^{2} + y^{2} = 8$ touch each other.

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Solution

Given equation of curves are:
$x y = 4 a n d x^{2} + y^{2} = 8 a n d 2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} a n d \frac{d y}{d x} = \frac{- 2 x}{2 y} \Rightarrow \frac{d y}{d x} = \frac{- y}{x} = m_{1} a n d \frac{d y}{d x} = \frac{- x}{y} = m_{2}$
Since both the curves have same slope,
$∴ \frac{- y}{x} = \frac{- x}{y} \Rightarrow - y^{2} = - x^{2} \Rightarrow x^{2} = y^{2}$
Using the values of $x^{2}$ in Eq. (ii), we get:
$y^{2} + y^{2} = 8 \Rightarrow y^{2} = 4 \Rightarrow y = \pm 2$
For $y = 2, x = \frac{4}{2} = 2$
and for $y = - 2, x = \frac{4}{- 2} = - 2$
Thus, the required points of intersection are (2, 2) and (-2,-2).
For (2, 2), $m_{1} = \frac{- y}{x} = \frac{- 2}{2} = - 1 a n d m_{2} = \frac{- x}{y} = \frac{- 2}{2} = - 1 ∵ m_{1} = m_{2} f o r (- 2, 2) m_{1} = \frac{- y}{x} = \frac{- (- 2)}{- 2} = - 1 a n d m_{2} = \frac{- x}{y} = \frac{- (- 2)}{- 2} = - 1$
Thus, for both the intersection points, we see that slope of both the curves are same. Hence, the curves intersect each other.

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