Prove that the curves `x y=4` and `x^2+y^2=8` touch each other.

Answer

These two curves touch each other at their point of intersections.

Explanation

ATQ,

Given that, curves `xy = 4` .... (i) And `x^2 + y^2 = 8` ..... (ii)


The point of intersection of curves would be

From equation (i), `y=\frac{4}{x}`

Putting this in equation (ii), we would get

`\begin{array}{ll} x^{2}+\frac{16}{x^{2}}=8 \\ \Rightarrow x^{4}-8 x^{2}+16=0 \\ \Rightarrow (x^{2}-4)^{2}=0 \\ \Rightarrow x^{2}=4 \\ \Rightarrow x=\pm 2 \\ \Rightarrow y=\pm 2 \text { (using (i)) } \\ \therefore \text { Points of intersections }(2,2) \text { and }(-2,-2)\end{array}
For curve (i), `y+x(\frac{d y}{d x})_{1}=0`

At point `(2,2)` and `(-2,-2),(\frac{d y}{d x})_{1}=-1`

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