Question Number 192265 by yaslm last updated on 13/May/23
Answered by Frix last updated on 13/May/23
$${x}={p}+\mathrm{1} \\ $$$${y}={q}+\mathrm{1} \\ $$$$\left({p},\:{q}\right)\:\rightarrow\:\left(\mathrm{0},\:\mathrm{0}\right) \\ $$$$\frac{\left({x}−{y}\right)^{\mathrm{2}} }{{x}−{y}^{\mathrm{2}} }=−\frac{\left({p}−{q}\right)^{\mathrm{2}} }{{q}^{\mathrm{2}} −{p}+\mathrm{2}{q}} \\ $$$${p}={r}\mathrm{cos}\:\theta \\ $$$${q}={r}\mathrm{sin}\:\theta \\ $$$${r}\rightarrow\mathrm{0} \\ $$$$\frac{\left({p}−{q}\right)^{\mathrm{2}} }{{q}^{\mathrm{2}} −{p}+\mathrm{2}{q}}=\frac{\left(\mathrm{1}−\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\right){r}}{{r}\mathrm{sin}^{\mathrm{2}} \:\theta\:+\mathrm{2sin}\:\theta\:−\mathrm{cos}\:\theta} \\ $$$$\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{0} \\ $$