Question

Given two parabolas y=8-3x-2x^2 , y=2+9x-2x^2. Find values on parameters a and b so that straight y=ax+b touches both of the parabolas. Also find the coordinates where it touches the parabolas.

Answer 1

Of*

Answer 2

y=8-3x-2x^2 , y=2+9x-2x^2


Y = ax+b

8-3x-2x^2 = ax+b (1)

2+9x-2x^2 = ax+b (2)

(1) -> 2x^2 + (3+a)x + (b-8) = 0

(2) -> 2x^2 + (a-9) + (b-2) = 0

i think we are going to use the discriminant and form equations in just a and b

Answer 3

Does y=ax+b need to be tangent to parabolas ?

Answer 4

correction:
(2) -> 2x^2 + (a-9)x + (b-2) = 0

Answer 5

i think we do need it as tangent or there are a lot of solutions

Answer 6

But your solution does not indicate that.

Answer 7

can we use calculus?

Answer 8

Yes.

Answer 9

(1) -> 2x^2 + (3+a)x + (b-8) = 0

(2) -> 2x^2 + (a-9)x + (b-2) = 0

for quad eq. px^2 + qx + r = 0 we have D = q^2 - 4pr

discriminants:

3) -> (3+a)^2 - 8(b-8) = 0

4) -> (a-9)^2 -8(b-2)


my solution does, D = 0 implies tangent

Answer 10

@TuringTest Please do use calculus :) I know you can solve it

Answer 11

then just take the derivative and find when the tangents are of equal slope
proceed from there

Answer 12

ah sorry i didnt realise we wanted calc :)

Answer 13

But how do we know that they do not intersect?

Answer 14

@alireza.safdari

is that a question to me?

Answer 15

because the value of the slopes should be the same, which means the lines are parallel
parallel lines do not intersect
I think I may have oversimplified something though, hold on...

Answer 16

:D i'll leave it to you guys, im tired

Answer 17

let's take an easier example y=x^2 and Y= (x+1)^2
y' = 2x and Y' = 2(x+1)
2x = 2x + 2, we can't continue from here, am I right?

Answer 18

The answer for example above is Y=0

Answer 19

yes I found the same problem here
it looks like there is no line that is tangent to both parabolas

Answer 20

There should be one. just connect the vertexes together

Answer 21

hm... what is the method to see that y=0 is the line?
it's clearly true for the last example, but how to show it....

Answer 22

It has just one intersection with both parabolas

Answer 23

yes, but how to show that mathematically
that would help us with this problem I think

Answer 24

An then the tangent of y=0 is equal to zero as it is zero in both parabolas at the point of intersection

Answer 25

I thought you want to know how to test it, but I don't know how to construct a solution yet

Answer 26

yeah I can see it's true, I'm looking for the method
the derivatives of these functions are linearly dependent, so we need the one solution that actually touches the parabola

Answer 27

I think I got it.
we find the intersection between line and both parabolas and we make sure delta is equal to zero which indicates that there is one solution and therefore the line has one intersection with each parabola and then we find the differentiation of both parabolas should be equal to the slope of the line. I am going to solve in on paper and see if it works

Answer 28

I would say it is the line joining the vertices of both parabolas

Answer 29

But however no common tangent exists for both parabola and hence I would say that no line "touches" both the parabolas.

Answer 30

I think one does, I just can't find it

Answer 31

as alireza pointed out it is hard to show that the line y=0 is the solution to the same Q for y=x^2 and y=(1+x^2)

Answer 32

@TuringTest , I agree with your explanation. But x^2 terms cancel out while equating both slopes. How to find a way out of it ??

Answer 33

continue from what eigenschmeigen said
3==> 73 + a^2 + 6a - 8b = 0
4==> 97 + a^2 - 18a - 8b = 0
now we know that at every parabola the x coordinate of intersections are:
for first one: -(3+a)/4
for second one: -(a-9) / 4
Now the slope of parabolas should be same at this point and equal to a,
for first one:
4*(-(3+a)/4 ) -3 = a ===> -3-a-3 = a ===> 2a = -6
for second one
4*( -(a-9) / 4) + 9 = a ====> -a+9 +9 = a ====> 2a = 18
so I think such a line does not exist now I try for
the example I gave
y = x^2 and Y = (x+1)^2
x^2 -ax - b =0
x^2 + (2-a) x + 1-b=0

finding D = 0
first one
a^2 +4b =0
second one
(2-a)^2 - 4 (1-b) = 0 ===> 4 -2a + a^2 -4 + 4b = 0 ==>
a^2 - 2a + 4b =0


then we find x for intersection:
for y is -a/2
for Y is -(2-a)/2
now we check the slope
for y the slope is -a at intersection with the line
for Y it is -2+a +2 at its intersection with the line
so now we get two equivalent equation:
a=-a ====> so a=0
now that a =0 we try our D equations and we get
for first one:
a^2 +4b =0 ===> 4b =0 ===> b=0
for second:
a^2 - 2a + 4b =0 ===> 4b=0 ==> b=0

so we get the right answer y=0x+0 ===> y=0

so those parabolas don't have such a line

Answer 34

I can say there is no such a line for the parabolas provided by ErkoT

Answer 35

aww dangit I just meant to change a small part and erased it :S
well how are you so sure no line exists?

Answer 36

@dumbcow especially tricky calculus problem

Answer 37

Because there is no line which can intersect both parabolas only at one point and it has the same slope.
In my solution first we find all lines that intersect both parabolas at one point.
then I say that at that point the slope of tangent should be equal to slope of the line which is passing those points.
So logically I am not assuming anything which can reduce the answers. And if there is no solution for a set of equation we conclude there is no answer which can satisfy all those equation and if all equation are coming from a right and logical place therefore we can say that there is no answer.

Answer 38

line is y = x+10

http://www.wolframalpha.com/input/?i=plot+8-3x-2x^2%2C+2%2B9x-2x^2%2Cx%2B10+from+-4+to+4

Answer 39

@dumbcow ,any specific way to find line equation or it is a guess out of no where :P

Answer 40

. The answer is correct, although I have no idea how to get it:|

Answer 41

haha no i used fact that derivative yields slope of tangent line
set slopes equal for both parabolas and set tangent lines equal for both parabolas

Let f(x) = 8-3x-2x^2 , x_1 will denote x_value where its tangent
g(x) = 2+9x-2x^2 , x_2 wil denote x_value where line is tangent to this parabola

a = f'(x) = g'(x)
\[-3-4x_1 = 9-4x_2\]
equation of tangent line:
\[y-f(x_1) = f'(x_1)(x-x_1)\]
\[y-g(x_2) = g'(x_2)(x-x_2)\]
set y_intercepts equal
\[b =f(x_1) -f'(x_1)x_1 = g(x_2)-g'(x_2)x_2\]
\[2(x_1)^{2}+8 = 2(x_2)^{2}+2\]
Now solve system of equations for x_1 and x_2

Answer 42

Great approach @dumbcow :D :D :D I will remember this :D

Answer 43

Can you tell me what I am missing in my solution?

Answer 44

not sure, i will need to look over all the posts...i have to go though so i will try to later today

Answer 45

thanks

Answer 46

My equation for slope of the point that they intersect is wrong.
I will end up with a=a. so I will be stuck there. Anyway @dobcow thanks

Answer 47

@alireza.safdari , these equations will work
3==> 73 + a^2 + 6a - 8b = 0
4==> 97 + a^2 - 18a - 8b = 0

if you solve this system you will get
a=1
b=10

--> y = x+10

Answer 48

Thank you @dumbcow

Answer 49

@dumbcow can we conclude that if a line is passing a parabola at on point it is tangent to the parabola ?

Answer 50

I think I will ask it as a new question

Answer 51

if it only passes through 1 point, then yes it is tangent
and if the slope of the line equals the derivative of the function at that point

Answer 52

I knew that such a solution existed...
I knew @dumbcow could find it!
nice one ;)