Question

the axis parallel to y axis and the parabola passes through the points (4,5),(-2,11),(-4,21) , find the equation of parabola.

Answer 1

Therefore the required equation of the parabola is

$x^2-12x }+62=6y$

Step-by-step explanation:

Given that , the axis of the parabola parallel to the y-axis.

Let the equation of the parabola is

ax²+bx+c=y

The parabola passes through the points (4,5),(-2,11) and (-4,21)

Therefore the points will be satisfy the equation of parabola,

Putting x=4 and y=5 in the  equation parabola,

a.4²+b.4+c=5

⇒16a+4b+c=5...........(1)

Again putting x=-2 and y= 11  in the  equation parabola,

a(-2)²+b(-2)+c=11

⇒4a-2b+c=11............(2)

Again putting x=-4 and y= 21  in the  equation parabola,

a(-4)²+b(-4)+c=21

⇒16a-4b+c=21.........(3)

Equation (1)- equation(2)

∴16a+4b+c-(16a-4b+c)=5-21

⇒16a+4b+c-16a+4b-c=-16

⇒8b= -16

⇒b= -2

Putting b=-2 in eqation (1) and (2) we get

16a+4(-2)+c=5        and          4a-2(-2)+c=11      

⇒16a +c=13 ....(4)                  ⇒4a+c=15..........(5)

Equation (4) - equation (5)

16a+c-(4a+c)=13-15

⇒16a +c-4a-c=2

⇒12a =2

$\Rightarrow a =\frac{2}{12}$

$\Rightarrow a =\frac{1}{6}$

Putting the value in equation (4) we get,

$16\times \frac{1}{6} +c=13$

$\Rightarrow \frac{8}{3} +c=13$

$\Rightarrow c=13-\frac{8}{3}$

$\Rightarrow c=\frac{31}{3}$

Therefore the required equation of the parabola is

$\frac{1}{6}x^2+(-2)x +\frac{31}{3}=y$

$\Rightarrow x^2-12x }+62=6y$