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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestBoth 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.

Log in to reply

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

Log in to reply

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.

Log in to reply

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)):

\(f(x)=x^{2}-4\)

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

Here it is :

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

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

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

Let's take a look at triangle \(ABC\) :

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

Let's draw circumcircle \(O3\)(orange) of the triangle \(ABC\) and find it's radius : \(R=\frac { AB*BC*CA }{ 4{ S }_{ ABC } } \), where \(R= 2.5\) .

So, the centre of \(O3\) is obviously point \((0;-1.5)\)

/

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

To prove that point \((0;-1.5)\) is the centre of \(O1\) we must prove that line segment \(LO1=1.5\).

Straight line \(LO1\) is a \(normal\)- line that is perpendicular to our parabola in point \(L\).

So, here it's function : \(f(x)=-\frac { 1 }{ 2{ x }_{ 0 } } (x-{ x }_{ 0 })+({ { x }_{ 0 } }^{ 2 }-4)\) (\({ x }_{ 0 } \) is the touch abscissa(point \(L\)) of the parabola and circle \(O1\))

Let's find \({ x }_{ 0 } \). Under our assumption point \(O1\) has the following coordinates \((0;-1.5)\). Now we must put them in our \(normal\) function.

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

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

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

Now, we should check if \(|LO1|=|O1D|=1.5\)

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

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

/

**********************************************************Now, the next part of my horrible solution.

It's much easier to find the centre of \(O2\).

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

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

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

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

Since \(KB\) and \(BD\) are tangents to the circle \(O2\) \(\Rightarrow \) \(KB=BD\). Also \(KO2=O2D\) \(\Rightarrow \) \(O2B\)-bisector and median \(\Rightarrow \) \(O2BD\approx { 52 }^{ \circ }\)

\(\tan { \angle } O2BD\approx 1.28\) \(\Rightarrow \) \(1.28=\frac { O2D }{ DB } =\frac { O2D }{ 2 }\) \(\Rightarrow \) \(O2D=2.5\)

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

/

**********************************************************And of course the last step :

\(\frac { { R }_{ O1 } }{ { R }_{ O2 } } =0.6\)

That's it \(=)\)

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply