Angle between tangents to an ellipse

The ellipse is given by:

$x^2 / a^2 + y^2 / b^2 = 1$

with $a = 2$ and $b = 1$ . The gradient at any point $(x, y)$ on the ellipse is given by:

$g = ( 2 x / a^2 , 2 y / b^2 )$

The given point from which the tangents are drawn is: $(x0, y0)$ , with $x0 = 4, y0 = 3$ .

Now, the condition for tangency is

$g . (x - x0, y - y0) = 0$

Hence

$2(x - x0) / a^2 + 2 y( y - y0) / b^2 = 0$

which, upon simplifying, becomes

$x x0 / a^2 + y y0 / b^2 = 1$

Now, we substitute this in ellipse equation:

$x^2 / a^2 + ( ( b^2 / y0 )^2 (1 - x x0 / a^2 )^2 / b^2 = 1$

Expanding,

$x^2 / a^2 + b^2 / y0^2 ( 1 - 2 x x0 / a^2 + x^2 x0^2 / a^4 ) = 1$

which is of the form,

$a_2 x^2 + a_1 x + a_0 = 0$

with

$a_2 = (1/a^2) + (x0^2 / y0^2 ) (b^2 / a^4) = (1/a^2) ( 1 + (x0 / y0)^2 (b/a)^2 )$

$a_1 = -2 (x0 / y0^2) (b/a)^2$

$a_0 = b^2 / y0^2 - 1$

So now, we can solve the quadratic equation for x, then solve for y from $x x0 / a^2 + y y0 / b^2 = 1$

Having found the two solution points (x1, y1) and (x2, y2) we can determine the gradients at each point as

$g1 = ( 2 x1 / a^2 , 2 y1 / b^2 )$

$g2 = ( 2 x2 / a^2 , 2 y2 / b^2 )$

Now, it is straight forward to find the angle between the two gradients, and to find the angle between the two tangents, we subtract this found angle from 180.

Substituting the given values of $a, b, x0, y0$ , we obtain the following tangency points:

$(-1.291, 0.764 )$ and $(1.907, -0.302 )$

From which the angle is computed as explained above, it comes to $\boxed{34.7}$ .