+118670 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 ?
+397 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
\(\displaystyle \mid\frac{64}{2(4-8)}\mid=8.\)
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.
+397 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
\(\displaystyle \mid\frac{64}{2(4-8)}\mid=8.\)
