Answer to Question #224017 in Analytic Geometry for Bless

Question

Show that the tangents from the point with position vector −2i − 3j to the ellipse 4x² + 9y² = 36 are perpendicular.

Answer 1

$$4x^2+9y^2=36$$

Differrentiate borh sides with respect to $x$

$$8x+18yy'=0$$

$$y'=-\dfrac{4x}{9y}$$

$$slope=m=-\dfrac{4x_0}{9y_0}$$

The equation of the tangent to the ellipse is

$$y-y_0=m(x-x_0)$$

$$y=mx-mx_0+y_0$$

$$x_0=-\dfrac{9y_0}{4}m$$

$$y_0-mx_0=\dfrac{y_0(4+9m^2)}{4}$$

$$4x_0^2+9y_0^2=36$$

$$\dfrac{81y_0^2}{4}m^2+9y_0^2=36$$

$$y_0^2(9m^2+4)=16$$

$$\dfrac{y_0}{4}=\pm\dfrac{1}{\sqrt{9m^2+4}}$$

$$y=mx\pm\sqrt{9m^2+4}$$

Point $(-2, -3)$

$$-3=m(-2)\pm\sqrt{9m^2+4}$$

$$2m-3=\pm\sqrt{9m^2+4}$$

$$4m^2-12m+9=9m^2+4$$

$$5m^2+12m-5=0$$

$$m_1m_2=\dfrac{-5}{5}=-1$$

Therefore the tangents from the point with position vector $-2i-3j$ to the ellipse $4x^2+9y^2=36$ are perpendicular.