+0  
 
-2
874
8
avatar+114910 

I have been wondering about this one

 

https://web2.0calc.com/questions/help-meh-pls_1

 

Maybe another mathematician might like to take a look ?

 Dec 1, 2020

Best Answer 

 #5
avatar+187 
+6

Algebra is a bit messy so needs checking, (and gaps filling in !).

 Result agrees with Melody's though.

 

The point P can be taken to have co-ordinates \(\displaystyle (\frac{t^{2}}{2},t)\).

( Parametric equations for the parabola can be \(\displaystyle x=t^{2}/2, y=t.\) )

 

Suppose that y = ax + b is the equation of a line passing through this point, then   \(\displaystyle t=a\frac{t^{2}}{2}+b,\)

so

\(\displaystyle y=ax+t-a\frac{t^{2}}{2}.\)   ................................................(1)

This cuts the circle \(\displaystyle x^{2}+y^{2}-2x=0\) at points with x co-ordinates given by (simplified eq),

\(\displaystyle x^{2}(1+a^{2})+x(2at-a^{2}t^{2}-2)+t^{2}+\frac{a^{2}t^{4}}{4}-at^{3}=0.\)

This needs to have equal roots if the contact with the circle is to be tangential, so

\(\displaystyle (2at - a^{2}t^{2}-2)^{2}-4(1+a^{2})(t^{2}+\frac{a^{2}t^{4}}{4}-a{t^3})=0,\)

or,

\(\displaystyle a^{2}(4t^{2}-t^{4})+a(4t^{3}-8t)+4-4t^{2}=0.\)

 

That gives us two values for a, (for the two lines from P to the circle).

Solving for a,

\(\displaystyle a_{1}=\frac{2(2+t-t^{2})}{t(4-t^{2})},\quad a_{2}=\frac{2(2-t-t^{2})}{t(4-t^{2})}.\)

Looking back at (1), the tangents will have intercepts with the y-axis at      \(\displaystyle t-a_{1}\frac{t^{2}}{2}\quad \text{ and }\quad t-a_{2}\frac{t^{2}}{2},\)

meaning that the base of the triangle PBC (the line between B and C along the y-axis), will have length  

\(\displaystyle (a_{1}-a_{2})\frac{t^{2}}{2}=\frac{2t^{2}}{4-t^{2}}, \\ \text{and the area of the triangle} \)

(base times height divided by 2),

\(\displaystyle \frac{2t^{2}}{4-t^{2}}.\frac{t^{2}}{2}.\frac{1}{2}=\frac{t^{4}}{2(4-t^{2})}. \)

(The denominator could be t^2 - 4, depending on whether t is less than or greater than 2)

 

Differentiating that in order to find max/min produces \(\displaystyle t=\pm\sqrt{8},\)

and an area

\(\displaystyle \mid\frac{64}{2(4-8)}\mid=8.\)

 Dec 1, 2020
 #1
avatar
0

I'll paste this problem in and maybe you look at the problem above?

 Dec 1, 2020
 #2
avatar
0

Let P be a variable point on the parabola y^2=2x and Let B and C be points on the y-axis so that the circle (x-1)^2+y^2=1 is inscribed in triangle PBC. Find the minimum area of triangle PBC.

 Dec 1, 2020
 #3
avatar+114910 
0

I am glad you are showing an interest but I want it answered on the original!

 

 

ANSWER ON THE ORIGINAL THREAD 

 

If you answer then you can leave a little note here stating that you have done so.

 Dec 1, 2020
edited by Melody  Dec 1, 2020
 #4
avatar
+1

👍 

.
 Dec 1, 2020
 #5
avatar+187 
+6
Best Answer

Algebra is a bit messy so needs checking, (and gaps filling in !).

 Result agrees with Melody's though.

 

The point P can be taken to have co-ordinates \(\displaystyle (\frac{t^{2}}{2},t)\).

( Parametric equations for the parabola can be \(\displaystyle x=t^{2}/2, y=t.\) )

 

Suppose that y = ax + b is the equation of a line passing through this point, then   \(\displaystyle t=a\frac{t^{2}}{2}+b,\)

so

\(\displaystyle y=ax+t-a\frac{t^{2}}{2}.\)   ................................................(1)

This cuts the circle \(\displaystyle x^{2}+y^{2}-2x=0\) at points with x co-ordinates given by (simplified eq),

\(\displaystyle x^{2}(1+a^{2})+x(2at-a^{2}t^{2}-2)+t^{2}+\frac{a^{2}t^{4}}{4}-at^{3}=0.\)

This needs to have equal roots if the contact with the circle is to be tangential, so

\(\displaystyle (2at - a^{2}t^{2}-2)^{2}-4(1+a^{2})(t^{2}+\frac{a^{2}t^{4}}{4}-a{t^3})=0,\)

or,

\(\displaystyle a^{2}(4t^{2}-t^{4})+a(4t^{3}-8t)+4-4t^{2}=0.\)

 

That gives us two values for a, (for the two lines from P to the circle).

Solving for a,

\(\displaystyle a_{1}=\frac{2(2+t-t^{2})}{t(4-t^{2})},\quad a_{2}=\frac{2(2-t-t^{2})}{t(4-t^{2})}.\)

Looking back at (1), the tangents will have intercepts with the y-axis at      \(\displaystyle t-a_{1}\frac{t^{2}}{2}\quad \text{ and }\quad t-a_{2}\frac{t^{2}}{2},\)

meaning that the base of the triangle PBC (the line between B and C along the y-axis), will have length  

\(\displaystyle (a_{1}-a_{2})\frac{t^{2}}{2}=\frac{2t^{2}}{4-t^{2}}, \\ \text{and the area of the triangle} \)

(base times height divided by 2),

\(\displaystyle \frac{2t^{2}}{4-t^{2}}.\frac{t^{2}}{2}.\frac{1}{2}=\frac{t^{4}}{2(4-t^{2})}. \)

(The denominator could be t^2 - 4, depending on whether t is less than or greater than 2)

 

Differentiating that in order to find max/min produces \(\displaystyle t=\pm\sqrt{8},\)

and an area

\(\displaystyle \mid\frac{64}{2(4-8)}\mid=8.\)

Tiggsy Dec 1, 2020
 #6
avatar+121047 
0

Thx, Tiggsy..  excellent work...that's a LOT to digest    !!!!

 

cool cool cool

CPhill  Dec 1, 2020
 #7
avatar
0

What do you mean 

 Dec 1, 2020
 #8
avatar+114910 
+1

Thanks Tiggsy,

That is great.

I will take a proper look a little later.

 Dec 1, 2020

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