Find the condition that the line …
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Sia ? 6 years, 5 months ago
we have, xy =1
{tex} \Rightarrow y = \frac{1}{x}{/tex}
{tex}\therefore\ \frac{dy}{dx}=-\frac{1}{x^2}{/tex}
The slope of the normal ={tex}x^2{/tex}
If ax+by+c=0 is normal to the curve xy=1,then
{tex}x^2=-\frac{a}{b} \ [\because slope\ of\ normal\ =-\frac{coeff.\ of\ x}{coeff. of\ y}]{/tex}
{tex}\therefore \; - \frac{a}{b} > 0{/tex}
{tex}\Rightarrow\ a>0,b<0 \ or\ a<0,b>0{/tex}
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