Find-the-equation-of-parabola-that-passes-through-points-1-2-3-4-and-tangents-to-the-line-y-x-25-3-

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 \to passes throught points

{\begin{cases} (1, 2) \to 2 = a + b + c \dots (i) \\ (3, 4) \to 4 = 9 a + 3 b + c \dots (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 \dots 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 \dots santuyy \dots

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 :

{\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