Let $M$ be an arbitrary point of an ellipse $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 $. We draw the perpendicular lines to its axis $MA, \; MB$ respectively. Prove that the pole of the line passing through the point $A, \; B$ lies on the curve $\displaystyle \frac{a^2}{x^2}+ \frac{b^2}{y^2}=1 $.
Solution:
Let $M(a\cot t, b \sin t), \; t \in [0, 2\pi)$ . Then $A(a \cos t, 0), \; B(0, b \sin t)$. If $P(x_0, y_0)$ the pole of the line passing through the points $A, \; B$ then:
$$ AB:=\frac{x x_0}{a_2}+ \frac{y y_0}{b^2}=1 $$
However $A, B \in AB$ hence $\displaystyle \frac{a x_0 \cos t}{a^2}=1 \Leftrightarrow \cos t = \frac{a}{x_0}$ and similarly $$ \frac{b y_0 \sin t}{b^2}=1 \Leftrightarrow \sin t = \frac{b}{y_0}$$
Since $$ \cos^2 t +\sin^2 t =1 \Leftrightarrow \frac{a^2}{x_0^2}+ \frac{b^2}{y_0^2}=1 $$ the result follows.
Solution:
Let $M(a\cot t, b \sin t), \; t \in [0, 2\pi)$ . Then $A(a \cos t, 0), \; B(0, b \sin t)$. If $P(x_0, y_0)$ the pole of the line passing through the points $A, \; B$ then:
$$ AB:=\frac{x x_0}{a_2}+ \frac{y y_0}{b^2}=1 $$
However $A, B \in AB$ hence $\displaystyle \frac{a x_0 \cos t}{a^2}=1 \Leftrightarrow \cos t = \frac{a}{x_0}$ and similarly $$ \frac{b y_0 \sin t}{b^2}=1 \Leftrightarrow \sin t = \frac{b}{y_0}$$
Since $$ \cos^2 t +\sin^2 t =1 \Leftrightarrow \frac{a^2}{x_0^2}+ \frac{b^2}{y_0^2}=1 $$ the result follows.
No comments:
Post a Comment