Question Number 51489 by peter frank last updated on 27/Dec/18
$${Given}\:{that}\:{y}={mx}+{c} \\ $$$${is}\:{equation}\:{of}\:\:{tangent} \\ $$$${to}\:{the}\:{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}\:} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${find}\:{coordinate}\:{of}\: \\ $$$${point}\:{of}\:{contact}. \\ $$
Answered by ajfour last updated on 27/Dec/18
$${P}\:\left({a}\mathrm{cos}\:\theta,{b}\mathrm{sin}\:\theta\right) \\ $$$$\frac{{x}\mathrm{cos}\:\theta}{{a}}+\frac{{y}\mathrm{sin}\:\theta}{{b}}−\mathrm{1}\:=\:\mathrm{0}\:\:\:\left({tangent}\right) \\ $$$${bx}\mathrm{cos}\:\theta+{ay}\mathrm{sin}\:\theta−{ab}\:=\:\mathrm{0} \\ $$$${or}\:\:\:\:\:{mx}−{y}+{c}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{{b}\mathrm{cos}\:\theta}{{m}}\:=\:\frac{{a}\mathrm{sin}\:\theta}{−\mathrm{1}}\:=\:\frac{−{ab}}{{c}} \\ $$$$\:\:\:\mathrm{tan}\:\theta\:=\:−\frac{{b}}{{am}}\:\: \\ $$$$\:\:\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} {m}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\:=\:\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\:\mid{c}\mid \\ $$$$\:\:{a}\mathrm{cos}\:\theta\:=\:{a}\left(\frac{−{am}}{\:\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right) \\ $$$$\:\:{b}\mathrm{sin}\:\theta\:=\:{b}\left(\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:{P}\:\left(−\frac{{a}^{\mathrm{2}} {m}}{{c}},\:\frac{{b}^{\mathrm{2}} }{{c}}\right)\:. \\ $$