+0

-10
1421
8
+118594

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

Maybe another mathematician might like to take a look ?

Dec 1, 2020

#5
+389
+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
0

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

Dec 1, 2020
#2
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
+118594
-1

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

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
+1

👍

.
Dec 1, 2020
#5
+389
+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

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

Tiggsy Dec 1, 2020
#6
+126971
0

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

CPhill  Dec 1, 2020
#7
0

What do you mean

Dec 1, 2020
#8
+118594
+1

Thanks Tiggsy,

That is great.

I will take a proper look a little later.

Dec 1, 2020