congruent? parabola

• Dec 26th 2006, 02:16 PM
shenton
congruent? parabola
Write an equation that defines each parabola.

vertex (1,2), opening up, congruent to y = 4x^2 - 5.

Answer key: y = 4(x-1)^2 + 2

Thanks.
• Dec 26th 2006, 09:21 PM
malaygoel
Quote:

Originally Posted by shenton
Write an equation that defines each parabola.

vertex (1,2), opening up, congruent to y = 4x^2 - 5.

Answer key: y = 4(x-1)^2 + 2

Thanks.

two parabola are congruent if they are same in evry respect, that is, you can superpose them.
Now, for two parabola, openint up in the seme direction, the thing that distinguishes them, as far as their shape is concerned is "a"(the distance between the focus and the vertex).
General equation of a parabola opening up is (x-h)^2 = 4a(y-k) where (h'k) is the vertex of the parabola.
In your original parabola(y = 4x^2 - 5), a is 1/4
So a in your parabola is going to be 1/16, since they ought to be congrent.
So, general equation is (x-1)^2=4*1/16(y-2)

Keep Smiling
Malay
• Dec 27th 2006, 04:42 AM
Soroban
Hello, shenton!

Quote:

Write an equation that defines each parabola.
vertex (1,2), opening up, congruent to y = 4x² - 5.

Answer key: y = 4(x-1)² + 2

If two similarly-oriented graphs are congruent,
. . one is a "translation" of the other.

The given parabola has its vertex at (0,-5).
The new parabola has its vertex at (1,2).

The parabola has been "moved" 1 unit to the right and 7 units up.

The new equation is: . y .= .4(x - 1)² - 5 + 7

Therefore: . y .= .4(x -1)² + 2

• Dec 27th 2006, 12:53 PM
shenton
Thanks for the detailed workings!

(Moderator: I now found this thread - I didn't know it was moved to pre-calculus - for that matter, I didn't know this is a pre-calculus topic. What I do know is this is a quardratic equation problem. Sorry for re-posting the question)