Given-that-point-P-a-cos-b-sin-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

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

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

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

$$c=ae=\sqrt{a^2-b^2}$$$$saypointN(u,v)$$$$eqn.ofPNis$$$$y=v-\frac{u-c}{v}(x-u)$$$$\Rightarrow (u-c)x+vy=u^2+v^2-uc$$$$sincePNistangenttoellipse,$$$$a^2{(u-c)}^2+b^2{v}^2={(u^2+v^2-uc)}^2{}^{\ast )}$$$$(u^2+v^2+a^2-b^2-2uc)(u^2+v^2-a^2)=0$$$$(u^2+v^2+c^2-2uc)(u^2+v^2-a^2)=0$$$$[{(u-c)}^2+v^2](u^2+v^2-a^2)=0$$$$\Rightarrow u^2+v^2-a^2=0$$$$i.e.thelocusofpointNis$$$$x^2+y^2-a^2=0$$$$or$$$$x^2+y^2=a^2$$$$$$$${}^{\ast )}seeQ157926$$

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

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