Ellipse Prove

[nycmath]

See attachment.

[Math Tutor]

Write the ellipse as [tex]x^2/a^2 + y^2/b^2 = 1[/tex], and put [tex]c^2 = a^2 - b^2, e = c/a[/tex].

The normal at P(x_1, y_1) is

[tex]\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2=c^2[/tex]

so it cuts the axes at

[tex]A\left(\frac{c^2x_1}{a^2},\,0\right),\qquad B\left(0,\,-\frac{c^2y_1}{b^2}\right)[/tex]

Hence

[tex]PA^2=\frac{b^4x_1^2}{a^4}+y_1^2,\qquad PB^2=x_1^2+\frac{a^4y_1^2}{b^4}[/tex]

Multiplying, the product is a perfect square:

[tex]PA^2\cdot PB^2=\left(\frac{b^2x_1^2}{a^2}+\frac{a^2y_1^2}{b^2}\right)^2\;\Longrightarrow\;PA\cdot PB=\frac{b^2x_1^2}{a^2}+\frac{a^2y_1^2}{b^2}[/tex]

Since P is on the ellipse, [tex]a^2y_1^2/b^2 = a^2 - x_1^2[/tex], so

[tex]PA\cdot PB=a^2-x_1^2+\frac{b^2x_1^2}{a^2}=a^2-\frac{c^2}{a^2}x_1^2=a^2-e^2x_1^2[/tex]

And the focal radii are [tex]F_1P = a + ex_1, F_2P = a - ex_1[/tex], so

[tex]F_1P\cdot F_2P=a^2-e^2x_1^2=PA\cdot PB[/tex]

which is what we had to prove.

[nycmath]

Write the ellipse as [tex]x^2/a^2 + y^2/b^2 = 1[/tex], and put [tex]c^2 = a^2 - b^2, e = c/a[/tex].

The normal at P(x_1, y_1) is

[tex]\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2=c^2[/tex]

so it cuts the axes at

[tex]A\left(\frac{c^2x_1}{a^2},\,0\right),\qquad B\left(0,\,-\frac{c^2y_1}{b^2}\right)[/tex]

Hence

[tex]PA^2=\frac{b^4x_1^2}{a^4}+y_1^2,\qquad PB^2=x_1^2+\frac{a^4y_1^2}{b^4}[/tex]

Multiplying, the product is a perfect square:

[tex]PA^2\cdot PB^2=\left(\frac{b^2x_1^2}{a^2}+\frac{a^2y_1^2}{b^2}\right)^2\;\Longrightarrow\;PA\cdot PB=\frac{b^2x_1^2}{a^2}+\frac{a^2y_1^2}{b^2}[/tex]

Since P is on the ellipse, [tex]a^2y_1^2/b^2 = a^2 - x_1^2[/tex], so

[tex]PA\cdot PB=a^2-x_1^2+\frac{b^2x_1^2}{a^2}=a^2-\frac{c^2}{a^2}x_1^2=a^2-e^2x_1^2[/tex]

And the focal radii are [tex]F_1P = a + ex_1, F_2P = a - ex_1[/tex], so

[tex]F_1P\cdot F_2P=a^2-e^2x_1^2=PA\cdot PB[/tex]

which is what we had to prove.

Perfect. Another great reply with valuable study information.