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Question-39469




Question Number 39469 by Raj Singh last updated on 06/Jul/18
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
let point be α ,β  calculation of slope is   (1/2)×(1/( (√x)))+(1/2)×(1/( (√y)))×(dy/dx)=0  (dy/dx)=−(((√y) )/( (√x)))  so tangent is y−β=−((√β)/( (√α)))(x−α)  to find op  put y=0  op=α+(√(αβ))   to find oQ putx=0  oQ=β+(√(αβ))  op+oQ=α+β+2(√(αβ))   again α ,β lies on (√x) +(√y) =(√a)  so (√α) +(√β) =(√a)  α+β+2(√(αβ)) =a proved
$${let}\:{point}\:{be}\:\alpha\:,\beta \\ $$$${calculation}\:{of}\:{slope}\:{is}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{{y}}}×\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\sqrt{{y}}\:}{\:\sqrt{{x}}} \\ $$$${so}\:{tangent}\:{is}\:{y}−\beta=−\frac{\sqrt{\beta}}{\:\sqrt{\alpha}}\left({x}−\alpha\right) \\ $$$${to}\:{find}\:{op}\:\:{put}\:{y}=\mathrm{0} \\ $$$${op}=\alpha+\sqrt{\alpha\beta}\: \\ $$$${to}\:{find}\:{oQ}\:{putx}=\mathrm{0} \\ $$$${oQ}=\beta+\sqrt{\alpha\beta} \\ $$$${op}+{oQ}=\alpha+\beta+\mathrm{2}\sqrt{\alpha\beta}\: \\ $$$${again}\:\alpha\:,\beta\:{lies}\:{on}\:\sqrt{{x}}\:+\sqrt{{y}}\:=\sqrt{{a}} \\ $$$${so}\:\sqrt{\alpha}\:+\sqrt{\beta}\:=\sqrt{{a}} \\ $$$$\alpha+\beta+\mathrm{2}\sqrt{\alpha\beta}\:={a}\:{proved} \\ $$$$ \\ $$

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