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Question Number 5615 by Rasheed Soomro last updated on 22/May/16

$$\mathrm{The}\mathrm{tangent}\mathrm{at}\mathrm{the}\mathrm{point}\mathrm{P}(a,b)\mathrm{on}\mathrm{the}$$$$\mathrm{curve}\mathrm{y}=\frac{ab}{x}\mathrm{meets}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{and}\mathrm{y}-\mathrm{axis}$$$$\mathrm{at}\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{respectively}.\mathrm{Show}\mathrm{that}$$$$\mathrm{PQ}=\mathrm{RP}.$$

Commented by prakash jain last updated on 22/May/16

$$\frac{dy}{dx}=-\frac{ab}{x^2}$$$$slopeoftangentat(a,b)=-\frac{ab}{a^2}=-\frac{b}{a}$$$$equationoftangent:y=mx+c$$$$y=-\frac{b}{a}x+c$$$$\mathrm{tangent}\mathrm{passes}\mathrm{thru}(a,b)$$$$b=-\frac{b}{a}a+c\Rightarrow c=2b$$$$y=-\frac{b}{a}x+2b$$$$x=0\Rightarrow y=2bor\mathrm{R}=(0,2b)$$$$y=0\Rightarrow x=2aor\mathrm{Q}=(2a,0)$$$$\mathrm{PQ}=\sqrt{{(2a-a)}^2+{(0-b)}^2}=\sqrt{a^2+b^2}$$$$\mathrm{QR}=\sqrt{{(0-a)}^2+{(2b-b)}^2}=\sqrt{a^2+b^2}$$$$\mathrm{PQ}=\mathrm{QR}$$

Commented by Rasheed Soomro last updated on 23/May/16

$$\mathcal{TH}\alpha n\mathcal{KS}!$$$$\mathrm{Always}\mathrm{miss}\mathrm{you}\mathrm{when}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{you}$$$$\mathrm{for}\mathrm{some}\mathrm{time}\mathrm{in}\mathrm{the}\mathrm{activity}\mathrm{of}\mathrm{the}\mathrm{forum}!$$

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