Find lengths of tangents drawn from $(3,-5)$ to the Ellipse Find lengths of tangents drawn from $A(3,-5)$ to the Ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
My try: I assumed the point of tangency of ellipse as $P(5\cos a, 4 \sin a)$ 
Now Equation of tangent at $P$ is given by
$$\frac{x \cos a}{5}+\frac{ y \sin a}{4}=1$$ whose slope is $$m_1=\frac{-4 \cot a}{5}$$
Also slope of $AP$ is given by
$$m_2=\frac{4 \sin a+5}{5 \cos a-3}$$
So both slopes are equal , with that we get
$$12 \cos a-25 \sin a=20 \tag{1}$$
Now distance $AP$ is given by
$$AP=\sqrt{(3-5 \cos a)^2+(5+4 \sin a)^2}=\sqrt{75-30 \cos a+40 \sin a} \tag{2}$$
Now using $(1)$ we have to find $\cos a$ and $\sin a$ and then substitute in $(2)$ which becomes lengthy.
Any better way?
 A: let $$y=mx+n$$ then the equation of the line through $P(3;-5)$ is given by
$$y=m(x-3)-5$$ plug this in the equation of the Ellipse and solve this equation for $x$, since it should be a tangentline, the discriminante must be Zero
you will get $$16m^2-30m-9=0$$ solve this equation for $m$
A: The slope of tangent at a point (x,y) on the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1$$ is  $$\frac {dy}{dx} = \frac {-16x}{25y}$$
On the other hand slope of the line passing through $(x,y)$ and $(3,-5)$ is $m = \frac {y+5}{x-3} $
$$  \frac {-16x}{25y}= \frac {y+5}{x-3} \implies 48x-125y=400 $$ 
Solve  the system $$ \frac{x^2}{25}+\frac{y^2}{16}=1$$  $$48x-125y=400 $$
to find  the two points $(-1.56199, -3.79980)$ and $(4.68293,-1.40176).$
Now we can find the distance from P to these points.
A: The polar of $(3,-5)$ w.r.t. the ellipse is $$\frac{3x}{25}+\frac{-5y}{16}-1=0.$$ Intersecting the polar and the ellipse you get the tangent points of the two tangents through $(3,-5).$
They are $(\frac{375\sqrt{41}+1200}{769},\frac{144\sqrt{41}-2000}{769})$ and $(\frac{-375\sqrt{41}+1200}{769},\frac{-144\sqrt{41}-2000}{769})$. The rest is just the distance between these points and $(3,-5)$ and the answer is the two lengths $$\frac{3}{769}\sqrt{1249475\pm 33210\sqrt{41}}.$$

Aside: If you want the equations of the tangents, you can look at the dual conic $25X^2+16Y^2-1=0.$ It intersects the dual line $3X-5Y+1=0$ in two points $$(-\frac{15\sqrt{41}+48}{769},-\frac{9\sqrt{41}-125}{769}),
 (\frac{15\sqrt{41}-48}{769},\frac{9\sqrt{41}+125}{769})$$ that have duals the tangent lines: $$-\frac{15\sqrt{41}+48}{769}x-\frac{9\sqrt{41}-125}{769}y+1=0$$ and $$\frac{15\sqrt{41}-48}{769}x+\frac{9\sqrt{41}+125}{769}y+1=0.$$ 

