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Question Number 116267 by bemath last updated on 02/Oct/20

Find the equation of parabola that passes through points (1,2) ,(3,4) and tangents to the line y=−x+((25)/3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{that} \\ $$$$\mathrm{passes}\:\mathrm{through}\:\mathrm{points}\:\left(\mathrm{1},\mathrm{2}\right)\:,\left(\mathrm{3},\mathrm{4}\right)\: \\ $$$$\mathrm{and}\:\mathrm{tangents}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=−\mathrm{x}+\frac{\mathrm{25}}{\mathrm{3}} \\ $$

Answered by bobhans last updated on 02/Oct/20

let the equation of parabola is y=ax^2 +bx+c →passes throught points { (((1,2)→2=a+b+c ...(i))),(((3,4)→4=9a+3b+c ...(ii))) :} substract eq (i) by (ii) ⇒2 = 8a+2b ; 1 = 4a+b ; b=1−4a since parabola tangent to the line y = −x+((25)/3), we get ax^2 +(1−4a)x+c=−x+((25)/3) ax^2 +(2−4a)x+c−((25)/3)= 0. Taking Δ = 4(1−2a)^2 −4a(c−((25)/3))=0 ⇒(1−2a)^2 = a(c−((25)/3)) ⇒ c = ((25)/3)+(((1−2a)^2 )/a); inserting to (i) ⇒2 = a+1−4a+((25)/3)+(((1−2a)^2 )/a) ⇒1 = ((25)/3)−3a+((1−4a+4a^2 )/a) ⇒−22=−9a+((3−12a+12a^2 )/a) ⇒−22a=−9a^2 +12a^2 −12a+3 ⇒3a^2 +10a+3 = 0 ⇒(3a+1)(a+3) = 0 for a=−(1/3)→ { ((b=1+(4/3)=(7/3))),((c=((25)/3)−3(1+(2/3))^2 = 0)) :} we get the equation of parabola is y= −(1/3)x^2 +(7/3)x for a=−3→ { ((b=13)),((c=((25)/3)−((49)/3)=−8)) :} we get the equation of parabola is y = −3x^2 +13x−8

$$\mathrm{let}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{is}\: \\ $$$$\mathrm{y}=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}\:\rightarrow\mathrm{passes}\:\mathrm{throught}\:\mathrm{points} \\ $$$$\begin{cases}{\left(\mathrm{1},\mathrm{2}\right)\rightarrow\mathrm{2}=\mathrm{a}+\mathrm{b}+\mathrm{c}\:...\left(\mathrm{i}\right)}\\{\left(\mathrm{3},\mathrm{4}\right)\rightarrow\mathrm{4}=\mathrm{9a}+\mathrm{3b}+\mathrm{c}\:...\left(\mathrm{ii}\right)}\end{cases} \\ $$$$\mathrm{substract}\:\mathrm{eq}\:\left(\mathrm{i}\right)\:\mathrm{by}\:\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\mathrm{2}\:=\:\mathrm{8a}+\mathrm{2b}\:;\:\mathrm{1}\:=\:\mathrm{4a}+\mathrm{b}\:;\:\mathrm{b}=\mathrm{1}−\mathrm{4a} \\ $$$$\mathrm{since}\:\mathrm{parabola}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\: \\ $$$$\mathrm{y}\:=\:−\mathrm{x}+\frac{\mathrm{25}}{\mathrm{3}},\:\mathrm{we}\:\mathrm{get}\:\mathrm{ax}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{4a}\right)\mathrm{x}+\mathrm{c}=−\mathrm{x}+\frac{\mathrm{25}}{\mathrm{3}} \\ $$$$\mathrm{ax}^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{4a}\right)\mathrm{x}+\mathrm{c}−\frac{\mathrm{25}}{\mathrm{3}}=\:\mathrm{0}.\:\mathrm{Taking} \\ $$$$\Delta\:=\:\mathrm{4}\left(\mathrm{1}−\mathrm{2a}\right)^{\mathrm{2}} −\mathrm{4a}\left(\mathrm{c}−\frac{\mathrm{25}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{2a}\right)^{\mathrm{2}} \:=\:\mathrm{a}\left(\mathrm{c}−\frac{\mathrm{25}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\:\mathrm{c}\:=\:\frac{\mathrm{25}}{\mathrm{3}}+\frac{\left(\mathrm{1}−\mathrm{2a}\right)^{\mathrm{2}} }{\mathrm{a}};\:\mathrm{inserting}\:\mathrm{to}\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\mathrm{2}\:=\:\mathrm{a}+\mathrm{1}−\mathrm{4a}+\frac{\mathrm{25}}{\mathrm{3}}+\frac{\left(\mathrm{1}−\mathrm{2a}\right)^{\mathrm{2}} }{\mathrm{a}} \\ $$$$\Rightarrow\mathrm{1}\:=\:\frac{\mathrm{25}}{\mathrm{3}}−\mathrm{3a}+\frac{\mathrm{1}−\mathrm{4a}+\mathrm{4a}^{\mathrm{2}} }{\mathrm{a}} \\ $$$$\Rightarrow−\mathrm{22}=−\mathrm{9a}+\frac{\mathrm{3}−\mathrm{12a}+\mathrm{12a}^{\mathrm{2}} }{\mathrm{a}} \\ $$$$\Rightarrow−\mathrm{22a}=−\mathrm{9a}^{\mathrm{2}} +\mathrm{12a}^{\mathrm{2}} −\mathrm{12a}+\mathrm{3} \\ $$$$\Rightarrow\mathrm{3a}^{\mathrm{2}} +\mathrm{10a}+\mathrm{3}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3a}+\mathrm{1}\right)\left(\mathrm{a}+\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{a}=−\frac{\mathrm{1}}{\mathrm{3}}\rightarrow\begin{cases}{\mathrm{b}=\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{3}}}\\{\mathrm{c}=\frac{\mathrm{25}}{\mathrm{3}}−\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} =\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{is} \\ $$$$\mathrm{y}=\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{3}}\mathrm{x} \\ $$$$\mathrm{for}\:\mathrm{a}=−\mathrm{3}\rightarrow\begin{cases}{\mathrm{b}=\mathrm{13}}\\{\mathrm{c}=\frac{\mathrm{25}}{\mathrm{3}}−\frac{\mathrm{49}}{\mathrm{3}}=−\mathrm{8}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{is} \\ $$$$\mathrm{y}\:=\:−\mathrm{3x}^{\mathrm{2}} +\mathrm{13x}−\mathrm{8}\: \\ $$$$ \\ $$

Commented by bemath last updated on 02/Oct/20

Commented by bemath last updated on 02/Oct/20

yes...correct

$$\mathrm{yes}...\mathrm{correct} \\ $$

Commented by mr W last updated on 02/Oct/20

generally there are infinitely many parabolas which fulfill the given conditions. there are two parabolas with axis in y direction. that′s what you got. there are two parabolas with axis in x direction. there are many other parabolas with skew axis.

$${generally}\:{there}\:{are}\:{infinitely}\:{many} \\ $$$${parabolas}\:{which}\:{fulfill}\:{the}\:{given} \\ $$$${conditions}. \\ $$$${there}\:{are}\:{two}\:{parabolas}\:{with}\:{axis} \\ $$$${in}\:{y}\:{direction}.\:{that}'{s}\:{what}\:{you}\:{got}. \\ $$$${there}\:{are}\:{two}\:{parabolas}\:{with}\:{axis} \\ $$$${in}\:{x}\:{direction}. \\ $$$${there}\:{are}\:{many}\:{other}\:{parabolas}\:{with} \\ $$$${skew}\:{axis}. \\ $$

Commented by mr W last updated on 02/Oct/20

Commented by bemath last updated on 02/Oct/20

yes...santuyy...

$$\mathrm{yes}...\mathrm{santuyy}... \\ $$

Answered by 1549442205PVT last updated on 03/Oct/20

Dente the equation of the parabol we need find by (P): y=ax^2 +bx+c.Since the P pass through A(1,2)and B(3,4) we have: { ((a+b+c=2(1))),((9a+3b+c=4(2))) :} Substract (1)from(2)we get 8a+2b=2⇒b=1−4a,replace into (1) we get a+1−4a+c=2⇒c=3a+1 ⇒(P):y=ax^2 +(1−4a)x+3a+1 Since P touches to y=−x+((25)/3), infer : ax^2 +(1−4a)x+3a+1=−x+25/3 ⇔3ax^2 +2(3−6a)x+9a−22=0 has unique root⇔Δ′=(3−6a)^2 −3a(9a−22) =0⇔9a^2 +30a+9=0⇔(3a+5)^2 =16 ⇔3a+5=±4⇔a∈{−3,−1/3) we obtain two parabol are: i)y=−3x^2 +13x−8 ii)y=−(1/3)x^2 +(7/3)x

$$\mathrm{Dente}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parabol}\:\mathrm{we} \\ $$$$\mathrm{need}\:\mathrm{find}\:\mathrm{by}\:\left(\mathrm{P}\right):\:\mathrm{y}=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}.\mathrm{Since} \\ $$$$\mathrm{the}\:\mathrm{P}\:\mathrm{pass}\:\mathrm{through}\:\mathrm{A}\left(\mathrm{1},\mathrm{2}\right)\mathrm{and}\:\mathrm{B}\left(\mathrm{3},\mathrm{4}\right) \\ $$$$\mathrm{we}\:\mathrm{have}: \\ $$$$\begin{cases}{\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{2}\left(\mathrm{1}\right)}\\{\mathrm{9a}+\mathrm{3b}+\mathrm{c}=\mathrm{4}\left(\mathrm{2}\right)}\end{cases} \\ $$$$\mathrm{Substract}\:\left(\mathrm{1}\right)\mathrm{from}\left(\mathrm{2}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{8a}+\mathrm{2b}=\mathrm{2}\Rightarrow\mathrm{b}=\mathrm{1}−\mathrm{4a},\mathrm{replace}\:\mathrm{into}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{a}+\mathrm{1}−\mathrm{4a}+\mathrm{c}=\mathrm{2}\Rightarrow\mathrm{c}=\mathrm{3a}+\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{P}\right):\mathrm{y}=\mathrm{ax}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{4a}\right)\mathrm{x}+\mathrm{3a}+\mathrm{1} \\ $$$$\mathrm{Since}\:\mathrm{P}\:\mathrm{touches}\:\mathrm{to}\:\mathrm{y}=−\mathrm{x}+\frac{\mathrm{25}}{\mathrm{3}},\:\mathrm{infer}\:: \\ $$$$\mathrm{ax}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{4a}\right)\mathrm{x}+\mathrm{3a}+\mathrm{1}=−\mathrm{x}+\mathrm{25}/\mathrm{3}\: \\ $$$$\Leftrightarrow\mathrm{3ax}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}−\mathrm{6a}\right)\mathrm{x}+\mathrm{9a}−\mathrm{22}=\mathrm{0}\:\mathrm{has} \\ $$$$\mathrm{unique}\:\mathrm{root}\Leftrightarrow\Delta'=\left(\mathrm{3}−\mathrm{6a}\right)^{\mathrm{2}} −\mathrm{3a}\left(\mathrm{9a}−\mathrm{22}\right) \\ $$$$=\mathrm{0}\Leftrightarrow\mathrm{9a}^{\mathrm{2}} +\mathrm{30a}+\mathrm{9}=\mathrm{0}\Leftrightarrow\left(\mathrm{3a}+\mathrm{5}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\Leftrightarrow\mathrm{3a}+\mathrm{5}=\pm\mathrm{4}\Leftrightarrow\mathrm{a}\in\left\{−\mathrm{3},−\mathrm{1}/\mathrm{3}\right) \\ $$$$\mathrm{we}\:\mathrm{obtain}\:\mathrm{two}\:\mathrm{parabol}\:\mathrm{are}: \\ $$$$\left.\mathrm{i}\right)\mathrm{y}=−\mathrm{3x}^{\mathrm{2}} +\mathrm{13x}−\mathrm{8} \\ $$$$\left.\mathrm{ii}\right)\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{3}}\mathrm{x} \\ $$

Commented by mr W last updated on 03/Oct/20

we can also let the parabola be x=ay^2 +by+c but generally the parabola is (ax+by)^2 +cx+dy+e=0

$${we}\:{can}\:{also}\:{let}\:{the}\:{parabola}\:{be} \\ $$$${x}={ay}^{\mathrm{2}} +{by}+{c} \\ $$$$ \\ $$$${but}\:{generally}\:{the}\:{parabola}\:{is} \\ $$$$\left({ax}+{by}\right)^{\mathrm{2}} +{cx}+{dy}+{e}=\mathrm{0} \\ $$

Commented by 1549442205PVT last updated on 07/Oct/20

Yes,then we obtain more two parabol which symetry to two above paraboles through line y=x

$$\mathrm{Yes},\mathrm{then}\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{more}\:\mathrm{two}\:\mathrm{parabol}\: \\ $$$$\mathrm{which}\:\mathrm{symetry}\:\mathrm{to}\:\mathrm{two}\:\mathrm{above}\:\mathrm{paraboles}\: \\ $$$$\mathrm{through}\:\mathrm{line}\:\mathrm{y}=\mathrm{x} \\ $$

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