Anyone help !!!

I found this problem was very difficult :(. Can anyone help me solve this? Here is the problem:

Given a parabola y=x28x+12y = x^2 - 8x + 12. Find the ratio of two inscribed circle on the parabola, both bordered by the parabola and the xx axis. But one circle is over xx axis and the other one is under xx axis.

Note by Muh Amin
6 years ago

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the length of the circumference O1 = 15.71(approximately)

the length of the circumference O2 = 9.43(approximately)

Co1/Co2 = 1.7

Calculations made with Geogebra

John Doe - 6 years ago

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Is there no other 'common' way to solve this problem? I mean without using software? I just want to know how to solve this problem. But firstly, thank you so much for your help.

Muh Amin - 5 years, 12 months ago

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I don't know English very well, but I'll try to help you :

First of all let's rewrite our function in this form (it will be the same result at the end since I am using Translation (geometry)):


Then we must find the coordinates of the vertex of our new parabola and roots of x24=0x^{2}-4=0

Here it is :

1) coordinates of the vertex : x=0 x=0 and y=4y=-4 ,so : (0;4)(0;-4)

2) roots : x=±2 x = \pm2 , also we have two points : (2;0)(-2;0) and (2;0)(2;0)

So, now we should define some points (take a look at my picture) : A(0;4) A (0;-4) , B(2;0) B (-2;0) , C(2;0) C (2;0) also let's suppose that O1O1 and O2O2 are the centres of our circles, where O1(red)(0;y) O1(red) (0;y) and O2(0;y1) O2 (0;y1) .

Let's take a look at triangle ABCABC :

AB=AC=20AB=AC = \sqrt { 20 }(by Pythagorean theorem) and SΔABC=8 { S }_{ \Delta ABC } = 8

Let's draw circumcircle O3O3(orange) of the triangle ABCABC and find it's radius : R=ABBCCA4SABCR=\frac { AB*BC*CA }{ 4{ S }_{ ABC } } , where R=2.5R= 2.5 .

So, the centre of O3O3 is obviously point (0;1.5)(0;-1.5)


Let's suppose that point (0;1.5)(0;-1.5) is also the centre of circle O1O1. It means, that R(O1)=1.5R(O1)=1.5.

To prove that point (0;1.5)(0;-1.5) is the centre of O1O1 we must prove that line segment LO1=1.5LO1=1.5.

Straight line LO1LO1 is a normalnormal- line that is perpendicular to our parabola in point LL.

So, here it's function : f(x)=12x0(xx0)+(x024)f(x)=-\frac { 1 }{ 2{ x }_{ 0 } } (x-{ x }_{ 0 })+({ { x }_{ 0 } }^{ 2 }-4) (x0{ x }_{ 0 } is the touch abscissa(point LL) of the parabola and circle O1O1)

Let's find x0{ x }_{ 0 } . Under our assumption point O1O1 has the following coordinates (0;1.5)(0;-1.5). Now we must put them in our normalnormal function.

12x0(0x0)+(x024)=1.5-\frac { 1 }{ 2{ x }_{ 0 } } (0-{ x }_{ 0 })+({ { x }_{ 0 } }^{ 2 }-4)=-1.5

Where x0=1.414213562{ x }_{ 0 }=-1.414213562. Now we must put x0{ x }_{ 0 } in the parabola function.

y=(1.414213562)24y={(-1.414213562) }^{ 2 }-4 , y=2y=-2. Ordinate of point L=2L=-2. So, L=(1.414213562;2)L=(-1.414213562;-2)

Now, we should check if LO1=O1D=1.5|LO1|=|O1D|=1.5

LO1=(0(1.414213562))2+(1.5(2))2=1.5|LO1|=\sqrt { { (0-(-1.414213562)) }^{ 2 }+{ (-1.5-(-2)) }^{ 2 } } = 1.5.

Our assumptions were confirmed =) . So point (0;1.5)(0;-1.5) is the centre of circle O1O1. And of course O1D=RO1=1.5|O1D|={ R }_{ O1 }=1.5


Now, the next part of my horrible solution.

It's much easier to find the centre of O2O2.

Let's assume that the straight line KBKB coincides with our parabola (y(0;6)y\in (0;6)) and tangent line to the parabola at point B(2;0)B(-2;0)

So, we can find tan(KBD)\tan { (\angle KBD) }

Firstly, let's find the tangent line function : it's f(x)=(x24)=2xf(x)'=({ x }^{ 2 }-4)'=2x then put abscissa of point BB in this function.

So, f(2)=4=tanKBDKBD104f'(-2)=-4=\tan { \angle } KBD \Rightarrow { \angle } KBD\approx { 104 }^{ \circ }

Since KBKB and BDBD are tangents to the circle O2O2 \Rightarrow KB=BDKB=BD. Also KO2=O2DKO2=O2D \Rightarrow O2BO2B-bisector and median \Rightarrow O2BD52O2BD\approx { 52 }^{ \circ }

tanO2BD1.28\tan { \angle } O2BD\approx 1.28 \Rightarrow 1.28=O2DDB=O2D21.28=\frac { O2D }{ DB } =\frac { O2D }{ 2 } \Rightarrow O2D=2.5O2D=2.5

O2=(0;2.5)O2=(0;2.5) \Rightarrow RO2=2.5{ R }_{ O2 }=2.5


And of course the last step :

RO1RO2=0.6\frac { { R }_{ O1 } }{ { R }_{ O2 } } =0.6

That's it =)=)

John Doe - 5 years, 12 months ago

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@John Doe Thank you very much Fatum. My english is not very good also :)

Muh Amin - 5 years, 12 months ago

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@Muh Amin But why exactly this problem, why did you want to solve it?

John Doe - 5 years, 12 months ago

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@John Doe I just randomly solve problem from my book. And from those problems, I just can't solve this one

Muh Amin - 5 years, 11 months ago

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@Muh Amin What book?

John Doe - 5 years, 11 months ago

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@John Doe Text book. I'm a student of high school

Muh Amin - 5 years, 11 months ago

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Both the circles touch X axis at (4,0) by symmetry.We can interpret this as follows -these two circles are the members of family of circles that pass through through the intersection of line y=0 and the circle with centre (4,0) and radius 0.We know that S+ku=0 represents family of circles passing through intersection of circle S=0 and line u=0 where k is arbitrary parameter.Now solve the equation of circle and parabola to obtain a quadratic equation in y.Since the parabola and circle touch each other set the discriminant of the quadratic as 0.This will give two values of k namely 3 and -5.The two values of k give us the two equations of the circles.

Indraneel Mukhopadhyaya - 5 years, 10 months ago

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