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

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

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

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

(dy/dx)=−((ab)/x^2 )  slope of tangent at (a,b)=−((ab)/a^2 )=−(b/a)  equation of tangent: y=mx+c  y=−(b/a)x+c  tangent passes thru (a,b)  b=−(b/a)a+c⇒c=2b  y=−(b/a)x+2b  x=0⇒y=2b  or R=(0,2b)  y=0⇒x=2a or Q=(2a,0)  PQ=(√((2a−a)^2 +(0−b)^2 ))=(√(a^2 +b^2 ))  QR=(√((0−a)^2 +(2b−b)^2 ))=(√(a^2 +b^2 ))  PQ=QR

$$\frac{{dy}}{{dx}}=−\frac{{ab}}{{x}^{\mathrm{2}} } \\ $$$${slope}\:{of}\:{tangent}\:{at}\:\left({a},{b}\right)=−\frac{{ab}}{{a}^{\mathrm{2}} }=−\frac{{b}}{{a}} \\ $$$${equation}\:{of}\:{tangent}:\:{y}={mx}+{c} \\ $$$${y}=−\frac{{b}}{{a}}{x}+{c} \\ $$$$\mathrm{tangent}\:\mathrm{passes}\:\mathrm{thru}\:\left({a},{b}\right) \\ $$$${b}=−\frac{{b}}{{a}}{a}+{c}\Rightarrow{c}=\mathrm{2}{b} \\ $$$${y}=−\frac{{b}}{{a}}{x}+\mathrm{2}{b} \\ $$$${x}=\mathrm{0}\Rightarrow{y}=\mathrm{2}{b}\:\:{or}\:\mathrm{R}=\left(\mathrm{0},\mathrm{2}{b}\right) \\ $$$${y}=\mathrm{0}\Rightarrow{x}=\mathrm{2}{a}\:{or}\:\mathrm{Q}=\left(\mathrm{2}{a},\mathrm{0}\right) \\ $$$$\mathrm{PQ}=\sqrt{\left(\mathrm{2}{a}−{a}\right)^{\mathrm{2}} +\left(\mathrm{0}−{b}\right)^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{QR}=\sqrt{\left(\mathrm{0}−{a}\right)^{\mathrm{2}} +\left(\mathrm{2}{b}−{b}\right)^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{PQ}=\mathrm{QR} \\ $$

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

THαnKS!  Always miss you when we don′t see you  for some time in the activity of the forum!

$$\mathcal{TH}\alpha{n}\mathcal{KS}! \\ $$$$\mathrm{Always}\:\mathrm{miss}\:\mathrm{you}\:\mathrm{when}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{you} \\ $$$$\mathrm{for}\:\mathrm{some}\:\mathrm{time}\:\mathrm{in}\:\mathrm{the}\:\mathrm{activity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{forum}! \\ $$

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