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Question-160333




Question Number 160333 by cortano last updated on 27/Nov/21
Commented by blackmamba last updated on 28/Nov/21
 (∗)r : y=mx+3  (∗∗)s: y=px+3   (∗∗∗) gradien =y′=m    8x−2yy′ = 0⇒y′=((4x)/y) ∧ m=((y−3)/x)   ⇒((4x)/y)=((y−3)/x) ; 4x^2 =y^2 −3y  ⇒36+y^2 =y^2 −3y , y=−12→ { ((x=3(√5))),((x=−3(√5))) :}   then point  { ((P(3(√5) ,−12))),((Q(−3(√5) ,−12))) :}  ⇒area of △PTQ=(1/2)×6(√5) ×15=45(√5)
$$\:\left(\ast\right){r}\::\:{y}={mx}+\mathrm{3} \\ $$$$\left(\ast\ast\right){s}:\:{y}={px}+\mathrm{3} \\ $$$$\:\left(\ast\ast\ast\right)\:{gradien}\:={y}'={m} \\ $$$$\:\:\mathrm{8}{x}−\mathrm{2}{yy}'\:=\:\mathrm{0}\Rightarrow{y}'=\frac{\mathrm{4}{x}}{{y}}\:\wedge\:{m}=\frac{{y}−\mathrm{3}}{{x}} \\ $$$$\:\Rightarrow\frac{\mathrm{4}{x}}{{y}}=\frac{{y}−\mathrm{3}}{{x}}\:;\:\mathrm{4}{x}^{\mathrm{2}} ={y}^{\mathrm{2}} −\mathrm{3}{y} \\ $$$$\Rightarrow\mathrm{36}+{y}^{\mathrm{2}} ={y}^{\mathrm{2}} −\mathrm{3}{y}\:,\:{y}=−\mathrm{12}\rightarrow\begin{cases}{{x}=\mathrm{3}\sqrt{\mathrm{5}}}\\{{x}=−\mathrm{3}\sqrt{\mathrm{5}}}\end{cases} \\ $$$$\:{then}\:{point}\:\begin{cases}{{P}\left(\mathrm{3}\sqrt{\mathrm{5}}\:,−\mathrm{12}\right)}\\{{Q}\left(−\mathrm{3}\sqrt{\mathrm{5}}\:,−\mathrm{12}\right)}\end{cases} \\ $$$$\Rightarrow{area}\:{of}\:\bigtriangleup{PTQ}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}\sqrt{\mathrm{5}}\:×\mathrm{15}=\mathrm{45}\sqrt{\mathrm{5}}\: \\ $$$$ \\ $$

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