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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

thedistancebetweenABandPmustbemax,sowe′relookingforthetangentparalleltoAB

Commented by ajfour last updated on 30/May/18

thanks sir, the required steps i shall post.

thankssir,therequiredstepsishallpost.

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(acosθ,bsinθ)slopeoftangentatPis=−bcosθasinθ=slopeofAB⇒bacotθ=−(q−k)(p−h)⇒tanθ=−b(p−h)a(q−k)xP=a2(q−k)a2(q−k)2+b2(p−h)2yP=−b2(p−h)a2(q−k)2+b2(p−h)2.

Commented by MrW3 last updated on 12/Jun/18

very good job!

verygoodjob!

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