منتديات الرياضيات العربية - عرض مشاركة واحدة

30-11-2009, 02:08 PM

Equation of the tangent T₁ at A :

$y - y_{0} = m(x-x_{0})$

where m is the slope and x₀ ,y₀ are the coordinates of A

Now , MA is perpendicular to T₁

$m_{MA}\times m_{T_{1}} = - 1$

$m_{MA}= \frac{y_{A}-y_{M}}{x_{A}-x_{M}}= \frac{3-2\sqrt{3}-3}{0-(-2)}=-\sqrt{3}$

$m_{T_{1}}=\frac{-1}{-\sqrt{3}}= \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

Equation of T₁

$y - y_{0} = m(x-x_{0})$

$y -(3-2\sqrt{3})= \frac{\sqrt{3}}{3}(x-0)\Rightarrow y=\frac{\sqrt{3}}{3}x + 3 - 2\sqrt{3}$

Same method for T₂ , the tangent at B

$y=\sqrt{3}x + 3 + 2\sqrt{3}$

For the point of intersection D , set y = y

$\frac{\sqrt{3}}{3}x + 3 - 2\sqrt{3}=\sqrt{3}x + 3 + 2\sqrt{3}$

Solving : x = -6

$y = 3 -4\sqrt{3}$

$D(-6,3 -4\sqrt{3})$