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Question-199522

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Question Number 199522 by cherokeesay last updated on 04/Nov/23
Answered by cortano12 last updated on 05/Nov/23
(g_1 ) ≡ y=ax+b (tangent line ) where a = (1/(4((1/4)))) = 1 and b = (1/4)−(1/8)=(1/8) ∴ y=x+(1/8) (h_1 )≡ y=px+q (normal line) where p=−(1/a)=−(1/1)=−1 and q = (1/4)+(1/8)=(3/8) ∴ y=−x+(3/8) (∗) normal cuts the parabola ⇒ (−x+(3/8))^2 = (1/2)x x^2 −(3/4)x+(9/(64))=(1/2)x x^2 −(5/4)x+(9/(64))=0 (x−(1/8))(x−(9/8))=0 the other point is R((9/8), −(3/4)) equation of tangent at R((9/8),−(3/4)) is y = rx+s , where r=(1/(4(−(3/4))))=−(1/3) and s = −(3/4)+(1/3).(9/8) =−(3/4)+(3/8)=−(3/8) ∴ y=−(1/3)x−(3/8) Let α is the angle between two tangent ⇒tan α = ∣((1−(−(1/3)))/(1+1.(−(1/3))))∣ tan α = 2 ⇒α = arctan (2)
$$\left(\mathrm{g}_{\mathrm{1}} \right)\:\equiv\:\mathrm{y}=\mathrm{ax}+\mathrm{b}\:\left(\mathrm{tangent}\:\mathrm{line}\:\right) \\ $$$$\:\mathrm{where}\:\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\:\mathrm{1} \\ $$$$\:\mathrm{and}\:\mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\:\:\therefore\:\mathrm{y}=\mathrm{x}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\:\left(\mathrm{h}_{\mathrm{1}} \right)\equiv\:\mathrm{y}=\mathrm{px}+\mathrm{q}\:\left(\mathrm{normal}\:\mathrm{line}\right) \\ $$$$\:\:\mathrm{where}\:\mathrm{p}=−\frac{\mathrm{1}}{\mathrm{a}}=−\frac{\mathrm{1}}{\mathrm{1}}=−\mathrm{1} \\ $$$$\:\:\mathrm{and}\:\mathrm{q}\:=\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\:\therefore\:\mathrm{y}=−\mathrm{x}+\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\:\left(\ast\right)\:\mathrm{normal}\:\mathrm{cuts}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\:\:\Rightarrow\:\left(−\mathrm{x}+\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x} \\ $$$$\:\:\:\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}\mathrm{x}+\frac{\mathrm{9}}{\mathrm{64}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x} \\ $$$$\:\:\:\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\mathrm{x}+\frac{\mathrm{9}}{\mathrm{64}}=\mathrm{0} \\ $$$$\:\:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{8}}\right)\left(\mathrm{x}−\frac{\mathrm{9}}{\mathrm{8}}\right)=\mathrm{0} \\ $$$$\:\:\mathrm{the}\:\mathrm{other}\:\mathrm{point}\:\mathrm{is}\:\mathrm{R}\left(\frac{\mathrm{9}}{\mathrm{8}},\:−\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\:\mathrm{equation}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{R}\left(\frac{\mathrm{9}}{\mathrm{8}},−\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\:\mathrm{is}\:\mathrm{y}\:=\:\mathrm{rx}+\mathrm{s}\:,\:\mathrm{where}\:\mathrm{r}=\frac{\mathrm{1}}{\mathrm{4}\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\mathrm{and}\:\mathrm{s}\:=\:−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{9}}{\mathrm{8}}\:=−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{8}}=−\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\:\:\therefore\:\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}−\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\:\mathrm{Let}\:\alpha\:\mathrm{is}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{two} \\ $$$$\:\mathrm{tangent}\:\Rightarrow\mathrm{tan}\:\alpha\:=\:\mid\frac{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}+\mathrm{1}.\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)}\mid \\ $$$$\:\mathrm{tan}\:\alpha\:=\:\mathrm{2}\:\Rightarrow\alpha\:=\:\mathrm{arctan}\:\left(\mathrm{2}\right)\: \\ $$
Commented by cherokeesay last updated on 05/Nov/23
thank you sir !
$${thank}\:{you}\:{sir}\:! \\ $$

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