Contract 8
Let the ellipse (E) have the canonical equation \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^) 2}}} = 1\) \((0 < b < a)\)
Consider a circle (C) with center O and radius a with the equation \({x^2} + {y^2} = {a^2}\)
Set points \(M\left( {x;y} \right) \in \left( E \right)\) and \({M_1}\left( {x;{y_1}} \right) \in \left ( C \right)\) such that \(y\) and \({y_1}\) always have the same sign (When M is different from the two vertices \({A_1},{A_2}\) of (E)) (Fig. ten)
a) From the canonical equation of the ellipse (E), calculate \({y^2}\) in terms of \({x^2}\)
From the equation of the circle (C), calculate \({y_1}^2\) in terms of \({x^2}\)
b) Calculate the ratio \(\frac{{HM}}{{H{M_1}}} = \frac{y}{{{y_1}}}\) to \(a,b\)
Detailed explanation:
a) We have \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \” Rightarrow {y^2} = {b^2}\left( {1 – \frac{{{x^2}}}{{{a^2}}}} \right) = \frac{{\left( {{a^2} – {x^2}} \right){b^2}}}{{{a^2}}}\)
Similarly, \({M_1}\left( {x;{y_1}} \right) \in \left( C \right)\) so \({x^2} + {y_1}^2 = {a^ 2} \Rightarrow {y_1}^2 = {a^2} – {x^2}\)
b) We have: \(\frac{{{y^2}}}{{{y_1}^2}} = \frac{{\frac{{\left( {{a^2} – {x^2) }} \right){b^2}}}{{{a^2}}}}}{{{a^2} – {x^2}}} = \frac{{{b^2}}} {{{a^2}}}\).
So \(\frac{{HM}}{{H{M_1}}} = \frac{y}{{{y_1}}} = \frac{b}{a}\), ie \({y_1} = \frac{a}{b}y\)
