Question

please solve this entrance exam problem q no 17 pg 6 http://www.fiitjee.com/down/sol/amaths06.pdf

Answer 1

Hello, Aadithya,

I've taken a look at this problem, and have some ideas for starting a possible solution, but have not actually solved the problem.

One of the keys to understanding this problem is understanding the meaning of "locus." What does that word mean in this context, when used in conjunction with the word "vertices"?

Hint: if we vary the value of a, will the vertices of the resulting parabolic graphs stay in one place, or will they move?

Have you tried applying the quadratic formula to this problem? You do, after all, have a quadratic equation here.

Answer 2

varying a does not change the vertex of the graph
but how did they find the vertex
what is the concept used

Answer 3

how did they represent vertex in terms of a

Answer 4

Your formula for this bunch of parabolas has third term -2a. This third term definitely affects the location of the vertex. So, if a changes, the vertex is going to move accordingly.

This problem is complex, so would take some time to discuss. I have several other students waiting for help.

What I'd suggest is that you look at the given function for y
and identify the coefficients A, B and C.

Then write an expression for in the form

\[x=\frac{ -B \pm \sqrt{B ^{2}-4*A*C} }{ 2A }\]

Answer 5

Once you've done that, try to combine the terms under the radical; in other words, simplify the discriminant as much as possible.

Once you've gotten that far, let me know.

Answer 6

ok

Answer 7

i will try my best

Answer 8

@mathmale i did not get this problem

Answer 9

I'd be glad to help you with this problem sometime today (not right now).

As always, I'd like to see what YOU have done, so that I'll know better where you're coming from and what you need to know/do to complete the solution of this problem.

Can you somehow photograph your work and upload the image to OpenStudy? Or could you scan your work and upload that scan?

Answer 10

i tried taking a^3/3 taken

Answer 11

then convert it into y=4ax^2 form
but i did not get the form

Answer 12

if you can tell me how to do it

Answer 13

then i know how to proceed

Answer 14

sorry y^2=4ax form

Answer 15

i did not understand the solution though

Answer 16

Please back up a bit and tell me specifically what your goal is (what you're trying to do RIGHT NOW).

Answer 17

yes

Answer 18

We need to review: y=ax^2 + bx + c. This is a quadratic function, and its roots can be found using the quadratic equation, right?

Answer 19

yes

Answer 20

According to the quadratic formula, the roots are \[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

Answer 21

yes i know this but how will we find the vertex locus

Answer 22

the roots i think will show the points of the curve

Answer 23

As you may remember, Aadithya, that x= -b/2a represents the x-coordinate of the vertex of the parabola. You can find the y-coordinate simply by substituting this x value back into the original quadratic. I'm certain that you know this already.