The final value of a is more logically decided by finding the y value 2a^2. The smaller a gives 0.81 and the larger gives 1.3. The value clearly has to be >1, so it is the larger one.
@@SyberMath It depends on what you mean by the term "the circle is tangent to the parabola". The number of solutions to the problem depends on it. You are suggested by the drawing, but the circle centered at (0, -1) with radius 1 and is tangent to the parobola at (0,0). The straight line tangent to the curve can cross the curve beyond the tangent point, which is also the case for curves described by third-degree polynomials. What if we allow that the tangency of the curves is "local" and that the curves have a common tangent line at the tangent point, and they can intersect beyond this point?
The origin of the two solutions can be seen at about 10:00 where you equating gradients 4a positive = (h-a)/(y-r) This can be true if both of the RHS terms are negative. This corresponds to the circle touching the parabola from the inside , a>0 but i think h
hi! about 12:28 i feel like at that point it would be easier to recognize that (1 - 2a^2)^2 and (2a^2 - 1)^2 are the same and that we could combine them instead of expanding seperately great video!
I think it may have been better to argue more rigorously why the bigger solution is the correct one. The way to do it to calculate the value of h, and then observe that the combination of the resulting values for a and h would have lead to some contradiction about the circle, such as the circle not being tangent to the x-axis, which is a piece of information given by the problem you are supposed to use.
Both solutions are valid. Second one is for the case, when the circle is tangent to the left “wing” of parabola and intersects the right “wing” in two points
Disagree. Nothing in the question says the circle must be to the right of the parabola. All 5 solutions are meaningful. Left outside, left inside, right outside, right inside, and dead centre (tangential at x=y=0).
I did not realize that I've done this before until after I was done shooting it! Well, it was a long time ago. 😁
"Did you hear that old math teachers never die?
They just lose some of their functions.
Are you getting old?
Are you losing some of your functions? 😁
The final value of a is more logically decided by finding the y value 2a^2. The smaller a gives 0.81 and the larger gives 1.3. The value clearly has to be >1, so it is the larger one.
Good thinking!
@@SyberMath It depends on what you mean by the term "the circle is tangent to the parabola". The number of solutions to the problem depends on it.
You are suggested by the drawing, but the circle centered at (0, -1) with radius 1 and is tangent to the parobola at (0,0).
The straight line tangent to the curve can cross the curve beyond the tangent point, which is also the case for curves described by third-degree polynomials.
What if we allow that the tangency of the curves is "local" and that the curves have a common tangent line at the tangent point, and they can intersect beyond this point?
The origin of the two solutions can be seen at about 10:00 where you equating gradients
4a positive = (h-a)/(y-r)
This can be true if both
of the RHS terms are negative. This corresponds to the circle touching the parabola from the inside , a>0 but i think h
Nice!
hi! about 12:28
i feel like at that point it would be easier to recognize that (1 - 2a^2)^2 and (2a^2 - 1)^2 are the same and that we could combine them instead of expanding seperately
great video!
Thank you and Good thinking!
How can find it the surface between the parabola and the circle??? ...
Can we be certain that a
I think the smaller value responds to the case that circle is tangent to parabola from inside.
I think it may have been better to argue more rigorously why the bigger solution is the correct one. The way to do it to calculate the value of h, and then observe that the combination of the resulting values for a and h would have lead to some contradiction about the circle, such as the circle not being tangent to the x-axis, which is a piece of information given by the problem you are supposed to use.
I agree! I guess I was trying to keep it simple here.
Not a contradiction, just a different case with h
Both solutions are valid. Second one is for the case, when the circle is tangent to the left “wing” of parabola and intersects the right “wing” in two points
Nice!
Also with a=0 and x=y=0.
5 valid solutions.
Nice Somewhere equivalence is lost but don't want to know where😉
Disagree. Nothing in the question says the circle must be to the right of the parabola. All 5 solutions are meaningful. Left outside, left inside, right outside, right inside, and dead centre (tangential at x=y=0).
A good problem,, but pretty easy for JEE aspirants 👍
Yes, true
@@SyberMath Just don't take it in negative way,,,, :)
I'm not praising JEE aspirants neither saying it's an easy Question