[Math] Find lengths of tangents drawn from $(3,-5)$ to the Ellipse

algebra-precalculusconic sectionsgeometrytangent line

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?

Best Answer

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 $m$

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