Question

Find equation of parabola if vertex at (−4, 2), axis parallel to the x-axis and the line x = 6− 4y is a tangent.

Answer 1

@jim_thompson5910 Help again please?

Answer 2

(y-k)^2 = 4a(x-h)

(y-2)^2 = 4a(x+4)

Answer 3

so far so good, just have to figure out what to do with that tangent

one sec

Answer 4

would be sub in (−4, 2) into tangent?
like to get
(-2, 5/2)?

Answer 5

that's the thing, the tangent isn't at (-4,2)

Answer 6

well then i have no idea what to do with the tangent :(

Answer 7

same here, but trying something out

Answer 8

well, im gonna take a shower now, brb

Answer 9

we got the same name O_O

Answer 10

does it say where the tangent line is tangent at? like where the tangent point is?

Answer 11

you would find it by equating

x = 6− 4y and whatever the formula is for the parabola

Answer 12

@jim_thompson5910

Answer 13

but does that guarantee that x = 6− 4y is a tangent?

Answer 14

If you know calculus, you can solve for dy/dx of the parabola and set that equal to the slope of the line.

you can then find x and y as a function of a, substitute those values into the parabola and solve for a.

Answer 15

how would you do that? we dont even know the parabola.

Answer 16

do you know calculus ?

Answer 17

yea

Answer 18

take the derivative with respect to x of
(y-2)^2 = 4a(x+4)

2(y-2) dy/dx = 4a
dy/dx = 2a/(y-2)

find the slope of the line x = 6− 4y
is: 4y= -x + 6, y = -1/4 x +6/4
m= -1/4

-1/4 = 2a/(y-2)

solve for y

use x = 6− 4y
to solve for x in terms of a

use those in
(y-2)^2 = 4a(x+4)

to find a

Answer 19

Here's a graph

Answer 20

so i got
x= 6-4y
y= -8a + 2

Answer 21

x= 6-4y
y= -8a + 2

so x= 6 -4(-8a+2) = 6 +32a-8 = 32a-2

use those in
(y-2)^2 = 4a(x+4)

Answer 22

how can we use those in (y-2)^2 = 4a(x+4)??

Answer 23

substitute y= -8a+2 and x= 32a-2 into (y-2)^2 = 4a(x+4)

Answer 24

ahh ok :)

Answer 25

Vertex: (-4,2)
Axis of Symmetry parallel to the x-axis
Tangent to x = 6-4y
This restricts the solution to one type. Opens in the negative x-direction.

Focus: (-4-a,2) for some a > 0
Directrix: x = -4 + a

Parabolas is all ordered pairs (x,y) such that the distance to the focus is the distance to the directrix.

Distance to Focus: \(\sqrt{(x+4+a)^{2}+(y-2)^{2}}\)

Distance to Directrix: x + 4 - a

Equating these and solving gives: \(-4a(x+4) = (y-2)^{2}\) Whix is good for \(x \le -4\)

Now for the line: x + 4y - 6 = 0

Distance to Directrix: x + 4 - a (Not very helpful, yet.)

Distance to Focus: \(\dfrac{|1(-4-a) + 4(2) - 6|}{\sqrt{1^{2}+4^{2}}} = \dfrac{|a+2|}{\sqrt{17}} = \dfrac{a+2}{\sqrt{17}}\), since a > 0, we can remove the absolute values.

The we know: \(x + 4 - a = \dfrac{a+2}{\sqrt{17}}\) This is the third required constraint.

Answer 26

Note: This is why we invented the calculus!