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


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

$Find the equation of parabola that$ $passes through points (1, 2), (3, 4)$ $and tangents to the line y = - x + \frac{25}{3}$
	Answered by bobhans last updated on 02/Oct/20
	$let the equation of parabola is y=ax^2 +bx+c →passes throught points { (((1,2)→2=a+b+c ...(i))),(((3,4)→4=9a+3b+c ...(ii))) :} substract eq (i) by (ii) ⇒2 = 8a+2b ; 1 = 4a+b ; b=1−4a since parabola tangent to the line y = −x+((25)/3), we get ax^2 +(1−4a)x+c=−x+((25)/3) ax^2 +(2−4a)x+c−((25)/3)= 0. Taking Δ = 4(1−2a)^2 −4a(c−((25)/3))=0 ⇒(1−2a)^2 = a(c−((25)/3)) ⇒ c = ((25)/3)+(((1−2a)^2 )/a); inserting to (i) ⇒2 = a+1−4a+((25)/3)+(((1−2a)^2 )/a) ⇒1 = ((25)/3)−3a+((1−4a+4a^2 )/a) ⇒−22=−9a+((3−12a+12a^2 )/a) ⇒−22a=−9a^2 +12a^2 −12a+3 ⇒3a^2 +10a+3 = 0 ⇒(3a+1)(a+3) = 0 for a=−(1/3)→ { ((b=1+(4/3)=(7/3))),((c=((25)/3)−3(1+(2/3))^2 = 0)) :} we get the equation of parabola is y= −(1/3)x^2 +(7/3)x for a=−3→ { ((b=13)),((c=((25)/3)−((49)/3)=−8)) :} we get the equation of parabola is y = −3x^2 +13x−8$
	$let the equation of parabola is$ $y = {ax}^{2} + bx + c \to passes throught points$ ${\begin{cases} (1, 2) \to 2 = a + b + c . . . (i) \\ (3, 4) \to 4 = 9 a + 3 b + c . . . (ii) \end{cases}$ $substract eq (i) by (ii)$ $\Rightarrow 2 = 8 a + 2 b; 1 = 4 a + b; b = 1 - 4 a$ $since parabola tangent to the line$ $y = - x + \frac{25}{3}, we get {ax}^{2} + (1 - 4 a) x + c = - x + \frac{25}{3}$ ${ax}^{2} + (2 - 4 a) x + c - \frac{25}{3} = 0 . Taking$ $Δ = 4 {(1 - 2 a)}^{2} - 4 a (c - \frac{25}{3}) = 0$ $\Rightarrow {(1 - 2 a)}^{2} = a (c - \frac{25}{3})$ $\Rightarrow c = \frac{25}{3} + \frac{{(1 - 2 a)}^{2}}{a}; inserting to (i)$ $\Rightarrow 2 = a + 1 - 4 a + \frac{25}{3} + \frac{{(1 - 2 a)}^{2}}{a}$ $\Rightarrow 1 = \frac{25}{3} - 3 a + \frac{1 - 4 a + {4 a}^{2}}{a}$ $\Rightarrow - 22 = - 9 a + \frac{3 - 12 a + {12 a}^{2}}{a}$ $\Rightarrow - 22 a = - {9 a}^{2} + {12 a}^{2} - 12 a + 3$ $\Rightarrow {3 a}^{2} + 10 a + 3 = 0$ $\Rightarrow (3 a + 1) (a + 3) = 0$ $for a = - \frac{1}{3} \to {\begin{cases} b = 1 + \frac{4}{3} = \frac{7}{3} \\ c = \frac{25}{3} - 3 {(1 + \frac{2}{3})}^{2} = 0 \end{cases}$ $we get the equation of parabola is$ $y = - \frac{1}{3} x^{2} + \frac{7}{3} x$ $for a = - 3 \to {\begin{cases} b = 13 \\ c = \frac{25}{3} - \frac{49}{3} = - 8 \end{cases}$ $we get the equation of parabola is$ $y = - {3 x}^{2} + 13 x - 8$
	Commented by bemath last updated on 02/Oct/20

	Commented by bemath last updated on 02/Oct/20

	$yes . . . correct$
	Commented by mr W last updated on 02/Oct/20

	$g e n e r a l l y t h e r e a r e i n f i n i t e l y m a n y$ $p a r a b o l a s w h i c h f u l f i l l t h e g i v e n$ $c o n d i t i o n s .$ $t h e r e a r e t w o p a r a b o l a s w i t h a x i s$ $i n y d i r e c t i o n . {t h a t}^{'} s w h a t y o u g o t .$ $t h e r e a r e t w o p a r a b o l a s w i t h a x i s$ $i n x d i r e c t i o n .$ $t h e r e a r e m a n y o t h e r p a r a b o l a s w i t h$ $s k e w a x i s .$
	Commented by mr W last updated on 02/Oct/20

	Commented by bemath last updated on 02/Oct/20

	$yes . . . santuyy . . .$
	Answered by 1549442205PVT last updated on 03/Oct/20
	$Dente the equation of the parabol we need find by (P): y=ax^2 +bx+c.Since the P pass through A(1,2)and B(3,4) we have: { ((a+b+c=2(1))),((9a+3b+c=4(2))) :} Substract (1)from(2)we get 8a+2b=2⇒b=1−4a,replace into (1) we get a+1−4a+c=2⇒c=3a+1 ⇒(P):y=ax^2 +(1−4a)x+3a+1 Since P touches to y=−x+((25)/3), infer : ax^2 +(1−4a)x+3a+1=−x+25/3 ⇔3ax^2 +2(3−6a)x+9a−22=0 has unique root⇔Δ′=(3−6a)^2 −3a(9a−22) =0⇔9a^2 +30a+9=0⇔(3a+5)^2 =16 ⇔3a+5=±4⇔a∈{−3,−1/3) we obtain two parabol are: i)y=−3x^2 +13x−8 ii)y=−(1/3)x^2 +(7/3)x$
	$Dente the equation of the parabol we$ $need find by (P) : y = {ax}^{2} + bx + c . Since$ $the P pass through A (1, 2) and B (3, 4)$ $we have :$ ${\begin{cases} a + b + c = 2 (1) \\ 9 a + 3 b + c = 4 (2) \end{cases}$ $Substract (1) from (2) we get$ $8 a + 2 b = 2 \Rightarrow b = 1 - 4 a, replace into (1)$ $we get a + 1 - 4 a + c = 2 \Rightarrow c = 3 a + 1$ $\Rightarrow (P) : y = {ax}^{2} + (1 - 4 a) x + 3 a + 1$ $Since P touches to y = - x + \frac{25}{3}, infer :$ ${ax}^{2} + (1 - 4 a) x + 3 a + 1 = - x + 25 / 3$ $\Leftrightarrow {3 ax}^{2} + 2 (3 - 6 a) x + 9 a - 22 = 0 has$ $unique root \Leftrightarrow Δ^{'} = {(3 - 6 a)}^{2} - 3 a (9 a - 22)$ $= 0 \Leftrightarrow {9 a}^{2} + 30 a + 9 = 0 \Leftrightarrow {(3 a + 5)}^{2} = 16$ $\Leftrightarrow 3 a + 5 = \pm 4 \Leftrightarrow a \in {- 3, - 1 / 3)$ $we obtain two parabol are :$ $i) y = - {3 x}^{2} + 13 x - 8$ $ii) y = - \frac{1}{3} x^{2} + \frac{7}{3} x$
	Commented by mr W last updated on 03/Oct/20

	$w e c a n a l s o l e t t h e p a r a b o l a b e$ $x = {a y}^{2} + b y + c$ $b u t g e n e r a l l y t h e p a r a b o l a i s$ ${(a x + b y)}^{2} + c x + d y + e = 0$
	Commented by 1549442205PVT last updated on 07/Oct/20

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