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Question Number 157867 by ZiYangLee last updated on 29/Oct/21

Given that point P(a cos θ, b sin θ) is a point on the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1. The tangent to the curve at point P is perpendicular to a straight line which passes through the focus, F (ae,0). If N is the intersection point, show that the equation of the locus of N is x^2 +y^2 =a^2 .

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{point}\:{P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}. \\ $$$$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{at}\:\mathrm{point}\:{P}\:\:\mathrm{is}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{focus}, \\ $$$${F}\:\left({ae},\mathrm{0}\right).\:\mathrm{If}\:{N}\:\mathrm{is}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{point},\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{N}\:\mathrm{is}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} . \\ $$

Answered by mr W last updated on 30/Oct/21

c=ae=(√(a^2 −b^2 )) say point N(u,v) eqn. of PN is y=v−((u−c)/v)(x−u) ⇒(u−c)x+vy=u^2 +v^2 −uc since PN is tangent to ellipse, a^2 (u−c)^2 +b^2 v^2 =(u^2 +v^2 −uc)^2 ^(∗)) (u^2 +v^2 +a^2 −b^2 −2uc)(u^2 +v^2 −a^2 )=0 (u^2 +v^2 +c^2 −2uc)(u^2 +v^2 −a^2 )=0 [(u−c)^2 +v^2 ](u^2 +v^2 −a^2 )=0 ⇒u^2 +v^2 −a^2 =0 i.e. the locus of point N is x^2 +y^2 −a^2 =0 or x^2 +y^2 =a^2 ^(∗)) see Q157926

$${c}={ae}=\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${say}\:{point}\:{N}\left({u},{v}\right) \\ $$$${eqn}.\:{of}\:{PN}\:{is} \\ $$$${y}={v}−\frac{{u}−{c}}{{v}}\left({x}−{u}\right) \\ $$$$\Rightarrow\left({u}−{c}\right){x}+{vy}={u}^{\mathrm{2}} +{v}^{\mathrm{2}} −{uc} \\ $$$${since}\:{PN}\:{is}\:{tangent}\:{to}\:{ellipse}, \\ $$$${a}^{\mathrm{2}} \left({u}−{c}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} {v}^{\mathrm{2}} =\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} −{uc}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:^{\left.\ast\right)} \\ $$$$\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{uc}\right)\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{uc}\right)\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left[\left({u}−{c}\right)^{\mathrm{2}} +{v}^{\mathrm{2}} \right]\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{2}} +{v}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${i}.{e}.\:{the}\:{locus}\:{of}\:{point}\:{N}\:{is} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${or} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$ \\ $$$$\:^{\left.\ast\right)} \:{see}\:{Q}\mathrm{157926} \\ $$

Commented by mr W last updated on 29/Oct/21

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