With the center at $(0,0)$ and $\psi=0$, the equation of the ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}$$
So $$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$$ and therefore $$\frac{dy}{dx}= -{\frac{xb^2}{ya^2}}\text{.}$$
The slope of the normal line is the negative reciprocal of this, so $$\tan(\phi)=\frac{ya^2}{xb^2}\text{.}$$
Meanwhile, $$\tan(\theta)=\frac{y}{x}\text{.}$$ So, eliminating $\frac{y}{x}$ from these two equations and clearing denominators, the relationship between $\phi$ and $\theta$ is: $$b^2\tan(\phi)=a^2\tan(\theta)$$
Kindly explain the reason behind this fact.
The reason is that an ellipse can be obtained by stretching/shrinking a circle. The strech/shrink is a linear map (linear transformation).
Let's consider two tangent lines on the circle $x^2+y^2=a^2$ at $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$. You already know that the two tangent lines are parallel.
Now, let's stretch/shrink the circle and the tangent lines. Stretching/shrinking the circle $x^2+y^2=a^2$ to obtain the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ means that you replace $y$ in $x^2+y^2=a^2$ with $\frac{a}{b}y$ to have $x^2+\left(\frac aby\right)^2=a^2$ which is nothing but $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
By this stretch/shrink, we have the followings :
The circle $x^2+y^2=a^2$ is transformed to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
The two parallel lines are transformed to two parallel lines.
The two lines tangent to the cirlce are transformed to two lines tangent to the ellipse.
The tangent points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$ on the circle are transformed to two tangent points $(a\cos\theta,b\sin\theta)$,$(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse respectively.
From the above facts, it follows that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$.
The followings are the proof for the above facts.
Let's consider the circle $x^2+y^2=a^2$ and two points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$.
The equation of the tangent line at $(a\cos\theta,a\sin\theta)$ is given by
$$a\cos\theta\ x+a\sin\theta\ y=a^2\tag1$$
Similarly, the equation of the tangent line at $(a\cos(\theta+\pi),a\sin(\theta+\pi))$ is given by
$$a\cos(\theta+\pi)x+a\sin(\theta+\pi)y=a^2\tag2$$
Now, let's stretch/shrink the circle and the lines $(1)(2)$ by replacing $y$ with $\frac aby$ to have
$$(1)\to a\cos\theta\ x+a\sin\theta\cdot\frac aby=a^2\tag3$$
$$(2)\to a\cos(\theta+\pi)x+a\sin(\theta+\pi)\cdot\frac aby=a^2\tag4 $$
Here note that these lines $(3)(4)$ are parallel since the slope of each line is $\frac{-b\cos\theta}{a\sin\theta}$.
Finally, note that $(3)$ can be written as
$$\frac{a\cos\theta}{a^2}x+\frac{b\sin\theta}{b^2}y=1\tag5$$
which is nothing but the tangent line at $(a\cos\theta,b\sin\theta)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Similarly, $(4)$ can be written as
$$\frac{a\cos(\theta+\pi)}{a^2}x+\frac{b\sin(\theta+\pi)}{b^2}y=1\tag6$$
which is nothing but the tangent line at $(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Since $(5)(6)$ are parallel, we see that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$. $\quad\square$
Best Answer
If the coordinates of the triangle are given by $(x_a,y_a)$, $(x_b,y_b)$ and $(x_c,y_c)$, then the area of the triangle is obtain as $$\left\vert\dfrac{x_a(y_b-y_c) + x_b(y_c-y_a) + x_c(y_a-y_b)}2 \right\vert$$ In your case, we obtain $$\dfrac{ab}2\left\vert \cos(t_a)(\sin(t_b)-\sin(t_c)) + \cos(t_b)(\sin(t_c)-\sin(t_a)) + \cos(t_c)(\sin(t_a)-\sin(t_b))\right\vert$$ Use trigonometric identities and simplify.