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

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

Answers (1)

Given: two curves xy = 4  and x^2+y^2 = 8

To prove: two curves meet each other at a point

Explanation:

Now given  x^2+y^2 = 8

Differentiating this with respect to x, we get

\begin{aligned} &\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(8)}{\mathrm{dx}}\\ &\text { By using the sum rule of differentiation, we get }\\ &\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0\\ &\Rightarrow 2 x+2 y \frac{d y}{d x}=0\\ &\Rightarrow 2 x=-2 y \frac{d y}{d x}\\ &\Rightarrow \frac{d y}{d x}=-\frac{x}{y}=m_{1} \ldots \ldots(i) \end{aligned}

 

Also given xy = 4

Differentiating this with respect to x, we get
\frac{\mathrm{d}(\mathrm{xy})}{\mathrm{dx}}=\frac{\mathrm{d}(4)}{\mathrm{dx}}$
The using the product rule of differentiation, we get

\Rightarrow\left(x \frac{d y}{d x}+y \frac{d x}{d x}\right)=0$ $\Rightarrow x \frac{d y}{d x}+y=0$
\\\Rightarrow \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}$\\ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{m}_{2} \ldots \ldots$ (ii)
But the touch of 2 curves is possible if
\mathrm{m}_{1}=\mathrm{m}_{2}$

Now substituting the values from equation (ii) and equation (ii), we get

\\-\frac{y}{x}=-\frac{x}{y}$\\ $\Rightarrow \mathrm{y}^{2}=\mathrm{x}^{2}$\\ $\Rightarrow \mathrm{x}=\mathrm{y}$
Now substituting x=y$ in $x^{2}+y^{2}=8,$ we get $y^{2}+y^{2}=8$
\\$\Rightarrow 2 \mathrm{y}^{2}=8$ \\$\Rightarrow y^{2}=4$ \\$\Rightarrow y=\pm 2$
When y=2$ $x y=4$ becomes $x(2)=4 \Rightarrow x=2$
when y=-2$ $x y=4$ becomes $$ x(-2)=4 \Rightarrow x=-2 $$

Therefore, (2,2) and (-2, -2) is the intersection point of the two curve

Substituting these points of intersection equation (i) and equation (ii), we get

For (2,2),

\\ \mathrm{m}_{1}=-\frac{\mathrm{x}}{\mathrm{y}}=-\frac{2}{2}=-1 \\ \mathrm{~m}_{2}=-\frac{\mathrm{y}}{\mathrm{x}}=-\frac{2}{2}=-1 \\ \therefore \mathrm{m}_{1}=\mathrm{m}_{2} \\ \text { For }(-2,-2) \\ \mathrm{m}_{1}=-\frac{\mathrm{x}}{\mathrm{y}}=-\frac{-2}{-2}=-1 \\ \mathrm{~m}_{2}=-\frac{\mathrm{y}}{\mathrm{x}}=-\frac{-2}{-2}=-1 \\ \therefore \mathrm{m}_{1}=\mathrm{m}_{2}

Thus, the condition for both the curves to touch is possible if that they have same slope

Hence the two given curves touch each other.

Hence proved

Posted by

infoexpert22

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