## Solution Section 6 Page 46 Math Learning Topic 10 – Kite>

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 $${y_1} = \frac{a}{b}y$$