# Thread: tangent with "unknown parabola"

1. ## tangent with "unknown parabola"

im sorry for posting so many questions. this is the last.

"the line $\displaystyle y=5x-3$ is a tangent to the parabola $\displaystyle y=ax^2 + bx -1$ at the point (1,2). find a and b"

now i have no idea where to start here. i could assume solving simultaneuosly could give some sort of answer but it would contain both unknowns.

if i differentiate $\displaystyle y=ax^2 + bx -1$ i get
$\displaystyle y'= 2ax+b$
and from the tangent we know that
$\displaystyle 2ax + b = 5$ at (1,2)

its here that im confused. whether the whole previous section is wrong i just cant get an answer.

im sorry for posting so many questions. this is the last.

"the line $\displaystyle y=5x-3$ is a tangent to the parabola $\displaystyle y=ax^2 + bx -1$ at the point (1,2). find a and b"

now i have no idea where to start here. i could assume solving simultaneuosly could give some sort of answer but it would contain both unknowns.

if i differentiate $\displaystyle y=ax^2 + bx -1$ i get
$\displaystyle y'= 2ax+b$
and from the tangent we know that
$\displaystyle 2ax + b = 5$ at (1,2)

its here that im confused. whether the whole previous section is wrong i just cant get an answer.
Just consider again what you are given, and use what you already have.
First, because the parabola must pass through the point (1,2) you must have that
$\displaystyle 2=a\cdot 1^2+b\cdot 1-1$
Second, since the given line with slope 5 is to be tangent to the parabola at point (1,2) you must also have that
$\displaystyle 5=2a\cdot 1+b$
In other words, you have a system of two linear equations for a and b. - Solve it. - Done!

3. $\displaystyle a\cdot 1^2+b\cdot 1-1 =2 (1)$
$\displaystyle 2a\cdot 1+b =5 (2)$
so simultaneously solved we get (2) - (1)
$\displaystyle a = 3$
and sub a=3 into (1)
$\displaystyle 3 + b -1 = 2$
$\displaystyle b-1 = -1$
$\displaystyle b=0$
is that right

therefore parabola eqn =
$\displaystyle y= 3x^2 -1$

maybe?

$\displaystyle a\cdot 1^2+b\cdot 1-1 =2 (1)$
$\displaystyle 2a\cdot 1+b =5 (2)$
so simultaneously solved we get (2) - (1)
$\displaystyle a = 3$
No, I think you get a=2
and sub a=3 into (1)
$\displaystyle 3 + b -1 = 2$
$\displaystyle b-1 = -1$
$\displaystyle b=0$
is that right
Not quite, you get b=1

therefore parabola eqn =
$\displaystyle y= 3x^2 -1$
maybe?
$\displaystyle y=2x^2+x-1$ would be more likely correct, I think.

5. oh yes 2 makes more sense. thanks a lot

6. Check the calcualtions.

a+b = 3...(1)

2a + b = 5 ...(2)

Now solve for a and b.