Question Number 184387 by ajfour last updated on 05/Jan/23
Commented by ajfour last updated on 05/Jan/23
$${If}\:{all}\:{three}\:{coloured}\:{areas}\:{are} \\ $$$${equal}\:{then}\:{find}\:{parameters}\: \\ $$$${h}\:{and}\:{k}\:{of}\:{parabola}\:{y}=\left({x}−{h}\right)^{\mathrm{2}} +{k}. \\ $$
Answered by mr W last updated on 06/Jan/23
Commented by mr W last updated on 06/Jan/23
Commented by mr W last updated on 07/Jan/23
![area of red triangle A_1 =((2×1)/2)=1 eqn. of line: y=(x/2)+1 eqn. of parabola: y=(x−h)^2 +k intersection of line and parabola: (x−h)^2 +k=(x/2)+1 (x−h)^2 −(((x−h))/2)+(k−1−(h/2))=0 ⇒x_2 −h=(1/4)+(1/4)(√(17+8h−16k)) ⇒y_2 =(9/8)+(h/2)+(1/8)(√(17+8h−16k)) area of blue region: A_3 =(1/6)[(1/4)−4(k−1−(h/2))]^(3/2) =A_1 =1 (1/6)[(1/4)−4(k−1−(h/2))]^(3/2) =1 17−16k+8h=4((36))^(1/3) ⇒k=(h/2)+((17)/(16))−(((36))^(1/3) /4) ⇒x_2 =h+(1/4)+((6)^(1/3) /2) ⇒y_2 =(h/2)+(9/8)+((6)^(1/3) /4) since A_2 =A_3 , area under the parabola is equal to area under the line from x=0 to x=x_2 . x_2 k+(h^3 /3)+(((x_2 −h)^3 )/3)=(((1+y_2 )x_2 )/2) (h+(1/4)+((6)^(1/3) /2))((h/2)+((17)/(16))−(((36))^(1/3) /4))+(h^3 /3)+((((1/4)+((6)^(1/3) /2))^3 )/3)=(((1+(h/2)+(9/8)+((6)^(1/3) /4))(h+(1/4)+((6)^(1/3) /2)))/2) ⇒h^3 +(3/4)(h+(1/4)+((6)^(1/3) /2))(h−((6)^(1/3) /2)−((36))^(1/3) )+((1/4)+((6)^(1/3) /2))^3 =0 since this is a cubic equation for h, exact solution is possible, but very complicated. we get approximately h≈1.5671, k≈1.0206](https://www.tinkutara.com/question/Q184406.png)
$${area}\:{of}\:{red}\:{triangle}\:{A}_{\mathrm{1}} =\frac{\mathrm{2}×\mathrm{1}}{\mathrm{2}}=\mathrm{1} \\ $$$${eqn}.\:{of}\:{line}: \\ $$$${y}=\frac{{x}}{\mathrm{2}}+\mathrm{1} \\ $$$${eqn}.\:{of}\:{parabola}: \\ $$$${y}=\left({x}−{h}\right)^{\mathrm{2}} +{k} \\ $$$${intersection}\:{of}\:{line}\:{and}\:{parabola}: \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +{k}=\frac{{x}}{\mathrm{2}}+\mathrm{1} \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} −\frac{\left({x}−{h}\right)}{\mathrm{2}}+\left({k}−\mathrm{1}−\frac{{h}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{2}} −{h}=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{17}+\mathrm{8}{h}−\mathrm{16}{k}} \\ $$$$\Rightarrow{y}_{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{8}}+\frac{{h}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}\sqrt{\mathrm{17}+\mathrm{8}{h}−\mathrm{16}{k}} \\ $$$${area}\:{of}\:{blue}\:{region}: \\ $$$${A}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{6}}\left[\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{4}\left({k}−\mathrm{1}−\frac{{h}}{\mathrm{2}}\right)\right]^{\mathrm{3}/\mathrm{2}} ={A}_{\mathrm{1}} =\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\left[\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{4}\left({k}−\mathrm{1}−\frac{{h}}{\mathrm{2}}\right)\right]^{\mathrm{3}/\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{17}−\mathrm{16}{k}+\mathrm{8}{h}=\mathrm{4}\sqrt[{\mathrm{3}}]{\mathrm{36}} \\ $$$$\Rightarrow{k}=\frac{{h}}{\mathrm{2}}+\frac{\mathrm{17}}{\mathrm{16}}−\frac{\sqrt[{\mathrm{3}}]{\mathrm{36}}}{\mathrm{4}} \\ $$$$\Rightarrow{x}_{\mathrm{2}} ={h}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}_{\mathrm{2}} =\frac{{h}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{8}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{4}} \\ $$$$ \\ $$$${since}\:{A}_{\mathrm{2}} ={A}_{\mathrm{3}} ,\:{area}\:{under}\:{the}\:{parabola} \\ $$$${is}\:{equal}\:{to}\:{area}\:{under}\:{the}\:{line}\:{from} \\ $$$${x}=\mathrm{0}\:{to}\:{x}={x}_{\mathrm{2}} . \\ $$$${x}_{\mathrm{2}} {k}+\frac{{h}^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({x}_{\mathrm{2}} −{h}\right)^{\mathrm{3}} }{\mathrm{3}}=\frac{\left(\mathrm{1}+{y}_{\mathrm{2}} \right){x}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left({h}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{2}}\right)\left(\frac{{h}}{\mathrm{2}}+\frac{\mathrm{17}}{\mathrm{16}}−\frac{\sqrt[{\mathrm{3}}]{\mathrm{36}}}{\mathrm{4}}\right)+\frac{{h}^{\mathrm{3}} }{\mathrm{3}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{3}}=\frac{\left(\mathrm{1}+\frac{{h}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{8}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{4}}\right)\left({h}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{h}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{4}}\left({h}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{2}}\right)\left({h}−\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{2}}−\sqrt[{\mathrm{3}}]{\mathrm{36}}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{6}}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$${since}\:{this}\:{is}\:{a}\:{cubic}\:{equation}\:{for}\:{h}, \\ $$$${exact}\:{solution}\:{is}\:{possible},\:{but}\:{very} \\ $$$${complicated}.\:{we}\:{get}\:{approximately} \\ $$$${h}\approx\mathrm{1}.\mathrm{5671},\:{k}\approx\mathrm{1}.\mathrm{0206} \\ $$
Commented by ajfour last updated on 06/Jan/23
$${Marvellous}\:{Sir}!\:{I}\:{shall}\:{attempt}\:{too}. \\ $$
Answered by ajfour last updated on 07/Jan/23
![If vertex of parabola is Origin. y=x^2 [eq. of parabola] y=(x/2)+(1+(h/2)−k) [eq. of line] say 1+(h/2)−k=m t^2 =(t/2)+m ∫_(−h) ^( t) (x^2 −(x/2)−m)dx=A (=0,1) ((t^3 /3)−(t^2 /4)+mt)+((h^3 /3)+(h^2 /4)+mh)=A ⇒ (t^3 /3)−(t^2 /4)+t^3 −(t^2 /2) +((h^3 /3)+(h^2 /4)+t^2 h−((th)/2))=A ⇒ (t+h){((t^2 −th+h^2 )/3)−(((t−h))/4) +t^2 −(t/2)}=A For A=0, t=t_2 as t+h≠0 ⇒ 4(t^2 −th+h^2 )+12t^2 +3h=6t+3t ⇒ 16t^2 −4th+4h^2 =9t−3h ⇒ 4((t^2 /h^2 )−(t/h)+1)=(3/h)(((3t)/h)−1) say (t/h)=s 4s^2 −(4+(9/h))s+(4+(3/h))=0 ⇒ s=(1/2)(1+(9/(4h)))±(√((1/4)(1+(9/(4h)))^2 −(1+(3/(4h))))) ⇒ 8hs=4h+9±(√((4h+9)^2 −16h(4h+3))) 8hs=4h+9±(√(81+24h−48h^2 )) ....](https://www.tinkutara.com/question/Q184465.png)
$${If}\:{vertex}\:{of}\:{parabola}\:{is}\:{Origin}. \\ $$$${y}={x}^{\mathrm{2}} \:\:\:\:\left[{eq}.\:{of}\:\:{parabola}\right] \\ $$$${y}=\frac{{x}}{\mathrm{2}}+\left(\mathrm{1}+\frac{{h}}{\mathrm{2}}−{k}\right)\:\:\:\:\:\left[{eq}.\:{of}\:\:{line}\right] \\ $$$${say}\:\:\:\mathrm{1}+\frac{{h}}{\mathrm{2}}−{k}={m} \\ $$$${t}^{\mathrm{2}} =\frac{{t}}{\mathrm{2}}+{m} \\ $$$$\int_{−{h}} ^{\:{t}} \left({x}^{\mathrm{2}} −\frac{{x}}{\mathrm{2}}−{m}\right){dx}={A}\:\:\:\left(=\mathrm{0},\mathrm{1}\right) \\ $$$$\left(\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}+{mt}\right)+\left(\frac{{h}^{\mathrm{3}} }{\mathrm{3}}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}+{mh}\right)={A} \\ $$$$\Rightarrow\:\:\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}+{t}^{\mathrm{3}} −\frac{{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\left(\frac{{h}^{\mathrm{3}} }{\mathrm{3}}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}+{t}^{\mathrm{2}} {h}−\frac{{th}}{\mathrm{2}}\right)={A} \\ $$$$\Rightarrow\:\left({t}+{h}\right)\left\{\frac{{t}^{\mathrm{2}} −{th}+{h}^{\mathrm{2}} }{\mathrm{3}}−\frac{\left({t}−{h}\right)}{\mathrm{4}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{t}^{\mathrm{2}} −\frac{{t}}{\mathrm{2}}\right\}={A} \\ $$$${For}\:{A}=\mathrm{0},\:\:{t}={t}_{\mathrm{2}} \\ $$$${as}\:\:{t}+{h}\neq\mathrm{0}\:\:\Rightarrow \\ $$$$\mathrm{4}\left({t}^{\mathrm{2}} −{th}+{h}^{\mathrm{2}} \right)+\mathrm{12}{t}^{\mathrm{2}} +\mathrm{3}{h}=\mathrm{6}{t}+\mathrm{3}{t} \\ $$$$\Rightarrow\:\:\:\mathrm{16}{t}^{\mathrm{2}} −\mathrm{4}{th}+\mathrm{4}{h}^{\mathrm{2}} =\mathrm{9}{t}−\mathrm{3}{h} \\ $$$$\Rightarrow\:\:\mathrm{4}\left(\frac{{t}^{\mathrm{2}} }{{h}^{\mathrm{2}} }−\frac{{t}}{{h}}+\mathrm{1}\right)=\frac{\mathrm{3}}{{h}}\left(\frac{\mathrm{3}{t}}{{h}}−\mathrm{1}\right) \\ $$$${say}\:\:\frac{{t}}{{h}}={s} \\ $$$$\mathrm{4}{s}^{\mathrm{2}} −\left(\mathrm{4}+\frac{\mathrm{9}}{{h}}\right){s}+\left(\mathrm{4}+\frac{\mathrm{3}}{{h}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{s}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}{h}}\right)\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}{h}}\right)^{\mathrm{2}} −\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}{h}}\right)} \\ $$$$\Rightarrow\:\mathrm{8}{hs}=\mathrm{4}{h}+\mathrm{9}\pm\sqrt{\left(\mathrm{4}{h}+\mathrm{9}\right)^{\mathrm{2}} −\mathrm{16}{h}\left(\mathrm{4}{h}+\mathrm{3}\right)} \\ $$$$\:\:\mathrm{8}{hs}=\mathrm{4}{h}+\mathrm{9}\pm\sqrt{\mathrm{81}+\mathrm{24}{h}−\mathrm{48}{h}^{\mathrm{2}} } \\ $$$$…. \\ $$