Question Number 36237 by ajfour last updated on 30/May/18
Answered by MJS last updated on 30/May/18
$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:{AB}\:\mathrm{and}\:{P}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{max},\:\mathrm{so}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{parallel} \\ $$$$\mathrm{to}\:{AB} \\ $$
Commented by ajfour last updated on 30/May/18
$${thanks}\:{sir},\:{the}\:{required}\:{steps}\:{i} \\ $$$${shall}\:{post}. \\ $$
Answered by ajfour last updated on 30/May/18
$${P}\left({a}\mathrm{cos}\:\theta,{b}\mathrm{sin}\:\theta\right) \\ $$$${slope}\:{of}\:{tangent}\:{at}\:{P}\:{is}\:\: \\ $$$$\:\:\:\:\:\:\:=−\frac{{b}\mathrm{cos}\:\theta}{{a}\mathrm{sin}\:\theta}\:=\:{slope}\:{of}\:{AB} \\ $$$$\Rightarrow\:\:\frac{{b}}{{a}}\mathrm{cot}\:\theta\:=\:−\frac{\left({q}−{k}\right)}{\left({p}−{h}\right)} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta\:=−\:\frac{{b}\left({p}−{h}\right)}{{a}\left({q}−{k}\right)} \\ $$$${x}_{{P}} =\frac{{a}^{\mathrm{2}} \left({q}−{k}\right)}{\:\sqrt{{a}^{\mathrm{2}} \left({q}−{k}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left({p}−{h}\right)^{\mathrm{2}} }} \\ $$$${y}_{{P}} =\frac{−{b}^{\mathrm{2}} \left({p}−{h}\right)}{\:\sqrt{{a}^{\mathrm{2}} \left({q}−{k}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left({p}−{h}\right)^{\mathrm{2}} }}\:\:. \\ $$$$ \\ $$
Commented by MrW3 last updated on 12/Jun/18
$${very}\:{good}\:{job}! \\ $$