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Question Number 36237 by ajfour last updated on 30/May/18

Answered by MJS last updated on 30/May/18

the distance between AB and P must be max, so we′re looking for the tangent parallel to AB

$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:{AB}\:\mathrm{and}\:{P}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{max},\:\mathrm{so}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{parallel} \\ $$$$\mathrm{to}\:{AB} \\ $$

Commented by ajfour last updated on 30/May/18

thanks sir, the required steps i shall post.

$${thanks}\:{sir},\:{the}\:{required}\:{steps}\:{i} \\ $$$${shall}\:{post}. \\ $$

Answered by ajfour last updated on 30/May/18

P(acos θ,bsin θ) slope of tangent at P is =−((bcos θ)/(asin θ)) = slope of AB ⇒ (b/a)cot θ = −(((q−k))/((p−h))) ⇒ tan θ =− ((b(p−h))/(a(q−k))) x_P =((a^2 (q−k))/(√(a^2 (q−k)^2 +b^2 (p−h)^2 ))) y_P =((−b^2 (p−h))/(√(a^2 (q−k)^2 +b^2 (p−h)^2 ))) .

$${P}\left({a}\mathrm{cos}\:\theta,{b}\mathrm{sin}\:\theta\right) \\ $$$${slope}\:{of}\:{tangent}\:{at}\:{P}\:{is}\:\: \\ $$$$\:\:\:\:\:\:\:=−\frac{{b}\mathrm{cos}\:\theta}{{a}\mathrm{sin}\:\theta}\:=\:{slope}\:{of}\:{AB} \\ $$$$\Rightarrow\:\:\frac{{b}}{{a}}\mathrm{cot}\:\theta\:=\:−\frac{\left({q}−{k}\right)}{\left({p}−{h}\right)} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta\:=−\:\frac{{b}\left({p}−{h}\right)}{{a}\left({q}−{k}\right)} \\ $$$${x}_{{P}} =\frac{{a}^{\mathrm{2}} \left({q}−{k}\right)}{\sqrt{{a}^{\mathrm{2}} \left({q}−{k}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left({p}−{h}\right)^{\mathrm{2}} }} \\ $$$${y}_{{P}} =\frac{−{b}^{\mathrm{2}} \left({p}−{h}\right)}{\sqrt{{a}^{\mathrm{2}} \left({q}−{k}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left({p}−{h}\right)^{\mathrm{2}} }}\:\:. \\ $$$$ \\ $$

Commented by MrW3 last updated on 12/Jun/18

very good job!

$${very}\:{good}\:{job}! \\ $$

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