Question Number 118482 by peter frank last updated on 17/Oct/20
$$\mathrm{If}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{focal}\:\:\mathrm{chord}\:\mathrm{of} \\ $$$$\mathrm{parabola}\:\mathrm{meet}\:\mathrm{the} \\ $$$$\mathrm{tangent}\:\mathrm{at}\:\mathrm{the}\:\:\mathrm{vertex} \\ $$$$\mathrm{in}\:\mathrm{C},\mathrm{D}.\mathrm{prove}\:\mathrm{that}\:\mathrm{CD} \\ $$$$\mathrm{substends}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angle} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{focus} \\ $$
Answered by mr W last updated on 18/Oct/20
![say vertex is the origin, the facus is F(0,h), then the eqn. of parabola is y=(x^2 /(4h)) say the focal chord is y=h+mx intersection with parabola: (x^2 /(4h))=h+mx x^2 −4mhx−4h^2 =0 ⇒x=2h(m±(√(m^2 +1))) end point A: x_A =2h(m−(√(m^2 +1))) y_A =h+2hm(m−(√(m^2 +1))) tangent at A: (dy/dx)=(x/(2h))=m−(√(m^2 +1)) y=h+2hm(m−(√(m^2 +1)))+(m−(√(m^2 +1)))[x−2h(m−(√(m^2 +1)))] 0=h+2hm(m−(√(m^2 +1)))+(m−(√(m^2 +1)))[x−2h(m−(√(m^2 +1)))] ⇒x=x_C =−(h/(m−(√(m^2 +1))))−2h(√(m^2 +1)) end point B: x_B =2h(m+(√(m^2 +1))) y_B =h+2hm(m+(√(m^2 +1))) tangent at B: y=h+2hm(m+(√(m^2 +1)))+(m+(√(m^2 +1)))[x−2h(m+(√(m^2 +1)))] 0=h+2hm(m+(√(m^2 +1)))+(m+(√(m^2 +1)))[x−2h(m+(√(m^2 +1)))] ⇒x=x_D =−(h/(m+(√(m^2 +1))))+2h(√(m^2 +1)) inclination of CF: m_(CF) =(h/(0−x_C ))=(1/((1/(m−(√(m^2 +1))))+2(√(m^2 +1)))) =((m−(√(m^2 +1)))/(1+2(√(m^2 +1))(m−(√(m^2 +1))))) =((m−(√(m^2 +1)))/(2m(√(m^2 +1))−2m^2 −1)) inclination of DF: m_(DF) =(h/(0−x_D ))=(1/((1/(m+(√(m^2 +1))))−2(√(m^2 +1)))) =((m+(√(m^2 +1)))/(1−2(√(m^2 +1))(m+(√(m^2 +1))))) =−((m+(√(m^2 +1)))/(2m(√(m^2 +1))+2m^2 +1)) m_(CF) ×m_(DF) =−((m−(√(m^2 +1)))/(2m(√(m^2 +1))−2m^2 −1))×((m+(√(m^2 +1)))/(2m(√(m^2 +1))+2m^2 +1)) =−((m^2 −(m^2 +1))/((2m(√(m^2 +1)))^2 −(2m^2 +1)^2 )) =−((−1)/(4m^4 +4m^2 −4m^4 −4m^2 −1)) =−1 ⇒CF⊥DF](Q118512.png)
$${say}\:{vertex}\:{is}\:{the}\:{origin},\:{the}\:{facus}\:{is} \\ $$$${F}\left(\mathrm{0},{h}\right),\:{then}\:{the}\:{eqn}.\:{of}\:{parabola}\:{is} \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}{h}} \\ $$$${say}\:{the}\:{focal}\:{chord}\:{is} \\ $$$${y}={h}+{mx} \\ $$$$ \\ $$$${intersection}\:{with}\:{parabola}: \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}{h}}={h}+{mx} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{mhx}−\mathrm{4}{h}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2}{h}\left({m}\pm\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$${end}\:{point}\:{A}: \\ $$$${x}_{{A}} =\mathrm{2}{h}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${y}_{{A}} ={h}+\mathrm{2}{hm}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${tangent}\:{at}\:{A}: \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}}{\mathrm{2}{h}}={m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$${y}={h}+\mathrm{2}{hm}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)+\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\left[{x}−\mathrm{2}{h}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\right] \\ $$$$\mathrm{0}={h}+\mathrm{2}{hm}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)+\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\left[{x}−\mathrm{2}{h}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\right] \\ $$$$\Rightarrow{x}={x}_{{C}} =−\frac{{h}}{{m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{2}{h}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$$ \\ $$$${end}\:{point}\:{B}: \\ $$$${x}_{{B}} =\mathrm{2}{h}\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${y}_{{B}} ={h}+\mathrm{2}{hm}\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${tangent}\:{at}\:{B}: \\ $$$${y}={h}+\mathrm{2}{hm}\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)+\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\left[{x}−\mathrm{2}{h}\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\right] \\ $$$$\mathrm{0}={h}+\mathrm{2}{hm}\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)+\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\left[{x}−\mathrm{2}{h}\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\right] \\ $$$$\Rightarrow{x}={x}_{{D}} =−\frac{{h}}{{m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}+\mathrm{2}{h}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$$ \\ $$$${inclination}\:{of}\:{CF}: \\ $$$${m}_{{CF}} =\frac{{h}}{\mathrm{0}−{x}_{{C}} }=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}+\mathrm{2}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\frac{{m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{1}+\mathrm{2}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)} \\ $$$$=\frac{{m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}{m}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}{m}^{\mathrm{2}} −\mathrm{1}} \\ $$$${inclination}\:{of}\:{DF}: \\ $$$${m}_{{DF}} =\frac{{h}}{\mathrm{0}−{x}_{{D}} }=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{2}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\frac{{m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{1}−\mathrm{2}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\left({m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)} \\ $$$$=−\frac{{m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}{m}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$$ \\ $$$${m}_{{CF}} ×{m}_{{DF}} =−\frac{{m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}{m}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}{m}^{\mathrm{2}} −\mathrm{1}}×\frac{{m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}{m}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=−\frac{{m}^{\mathrm{2}} −\left({m}^{\mathrm{2}} +\mathrm{1}\right)}{\left(\mathrm{2}{m}\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} −\left(\mathrm{2}{m}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{−\mathrm{1}}{\mathrm{4}{m}^{\mathrm{4}} +\mathrm{4}{m}^{\mathrm{2}} −\mathrm{4}{m}^{\mathrm{4}} −\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=−\mathrm{1} \\ $$$$\Rightarrow{CF}\bot{DF} \\ $$
Commented by peter frank last updated on 18/Oct/20
$$\mathrm{diagram}\:\mathrm{please} \\ $$
Commented by peter frank last updated on 18/Oct/20
$$\mathrm{thank}\:\mathrm{you}.\mathrm{i}\:\mathrm{appriaciate} \\ $$$$\mathrm{your}\:\mathrm{time}. \\ $$
Commented by mr W last updated on 18/Oct/20
Commented by peter frank last updated on 18/Oct/20
$$\mathrm{GODBLESS}\:\mathrm{YOU} \\ $$
Commented by peter frank last updated on 18/Oct/20
![where this came from y=h+2hm(m−(√(m^2 +1)))+(m−(√(m^2 +1)))[x−2h(m−(√(m^2 +1)))]](Q118596.png)
$$\mathrm{where}\:\mathrm{this}\:\mathrm{came}\:\mathrm{from} \\ $$$${y}={h}+\mathrm{2}{hm}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)+\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\left[{x}−\mathrm{2}{h}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\right] \\ $$
Commented by mr W last updated on 18/Oct/20
![point C(x_C ,y_C ) x_C =2h(m−(√(m^2 +1))) y_C =h+mx_C =h+2hm(m−(√(m^2 +1))) the inclination of tangent at C is m_C =m−(√(m^2 +1)) ⇒the eqn. of tangent at C is y=y_C +m_C (x−x_C ) ⇒y=h+2hm(m−(√(m^2 +1)))+(m−(√(m^2 +1)))[x−2h(m−(√(m^2 +1)))]](Q118622.png)
$${point}\:{C}\left({x}_{{C}} ,{y}_{{C}} \right) \\ $$$${x}_{{C}} =\mathrm{2}{h}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${y}_{{C}} ={h}+{mx}_{{C}} ={h}+\mathrm{2}{hm}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${the}\:{inclination}\:{of}\:{tangent}\:{at}\:{C}\:{is} \\ $$$${m}_{{C}} ={m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{the}\:{eqn}.\:{of}\:{tangent}\:{at}\:{C}\:{is} \\ $$$${y}={y}_{{C}} +{m}_{{C}} \left({x}−{x}_{{C}} \right) \\ $$$$\Rightarrow{y}={h}+\mathrm{2}{hm}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)+\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\left[{x}−\mathrm{2}{h}\left({m}−\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)\right] \\ $$