Consider an ellipse x^2/144 + y^2/64 = 1.A line is drawn tangent to the ellipse at a point P. A line segment drawn from the origin to a point Q on thi

Question

Consider an ellipse $x^2/144 + y^2/64 = 1$. A line is drawn tangent to the ellipse at a point P. A line segment drawn from the origin to a point Q on thi

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Answer 1

The equation of the given ellipse is

$$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $$

with $a = 12, b = 8$

The parametric equation of the ellipse is

$P = (x, y) = (a \cos t , b \sin t )$

The tangent vector at P is along the vector $V$,

$V = (- a \sin t, b \cos t )$

Therefore the equation of the tangent is

$q(t,s) = P + s V = (a \cos t , b \sin t ) + s (- a \sin t, b \cos t ) $

The minimum distance from the origin to $q(t)$ determines the point $Q$

$ q(t,s) \cdot q(t,s) = (P + s V) \cdot (P + s V) = P \cdot P + 2 P \cdot V + s^2 V \cdot V $

The minimum distance occurs at $ s_1 = - \dfrac{ P \cdot V}{ V \cdot V} $

This evaluates to:

$s_1 = \dfrac{(a^2 - b^2) \sin t \cos t }{ a^2 \sin^2 t + b^2 \cos^2 t } $

Therefore, point Q is given by

$Q = P + s_1 V = ( a \cos t , b \sin t ) + \dfrac{(a^2 - b^2) \sin t \cos t }{ a^2 \sin^2 t + b^2 \cos^2 t } (- a \sin t, b \cos t ) $

The area of $\triangle POQ $ is given by

$\text{Area} = \frac{1}{2} s_1 | P, V | = \frac{1}{2} s_1 (ab) $

Thus the maximum area corresponds to maximum $s_1$, so we want to maximize (over $t$), the following function:

$s1 = f(t) =\dfrac{(a^2 - b^2) \sin t \cos t }{ a^2 \sin^2 t + b^2 \cos^2 t } $

Using the double angle formula, this reduces to,

$f(t) = (a^2 - b^2) \dfrac{ \sin 2 t }{ (a^2 + b^2) + (b^2 - a^2) \cos 2 t } $

The numerator of $f'(t) / (a^2 - b^2) $ is given by

$g(t) = 2 \cos 2 t ( (a^2 + b^2) + (b^2 - a^2) \cos 2 t ) + 2 (b^2 - a^2) \sin^2 2 t $

And this reduces to

$g(t) = 2 \cos 2 t (a^2 + b^2) + 2 (b^2 - a^2 ) = 0$

Hence the solution is

$ \cos 2 t = \dfrac{a^2 - b^2}{a^2 + b^2} $

From which,

$ \sin 2 t = \sqrt{ 1 - \cos^2 2 t } = \dfrac{1}{a^2 + b^2} \sqrt{ (a^2 + b^2)^2 - (a^2 - b^2)^2} = \dfrac{2 a b}{a^2 + b^2}$

Using this value of $2 t$ we can evaluate $f(t)$ :

$f(t) = (a^2 - b^2) \dfrac{ \sin 2 t }{ (a^2 + b^2) + (b^2 - a^2) \cos 2 t } $

Substitute $\cos 2 t $ and $\sin 2 t $, to get,

$f(t) = (a^2 - b^2) (2 ab) \dfrac{1}{ (a^2 + b^2)^2 - (a^2 - b^2)^2 } = \dfrac{(a^2 - b^2)}{2 a b} $

Therfore the maximum area is

$ \text{Area Max } = \frac{1}{4} (a^2 - b^2)$

With $a^2 = 144 , b^2 = 64 $, we get

$ \text{Area Max } = \frac{1}{4} (144 - 64) = 20$