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WZ
Numerade Educator

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Problem 67 Hard Difficulty

Consider the tangent line to the ellipse $ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $ at a point $ (p, q) $ in the first quadrant.
(a) Show that the tangent line has $ x $-intercept $ a^2/p $ and $ y $-intercept $ b^2/q $.
(b) Show that the portion of the tangent line cut off by the coordinates axes has minimum length $ a + b $.
(c) Show that the triangle formed by the tangent line and the coordinate axes has minimum area $ ab $.

Answer

a)
$\begin{aligned} X-i n t e r c e p t &=\frac{a^{2}}{p} \\ Y-i n t e r c e p t &=\frac{b^{2}}{q} \end{aligned}$
b)
Minimum length $=a+b$
c)
Minimum Area formed is ab!

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Video Transcript

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If he is last on this number, have squired. Prom is greater than zero, so one P, if they got to a over square root of two, have square has a maximum value. It's a maximum value, a square. My small value is equal to the squire over a square hams. It's four over two minus to four Oh, for this is because you're a squire. Squire. Over four. So half max. Mon This culture it be over too, then a minimal. This is a half hams, Esquire, overpay square. I scored some space square over. Maybe over to this is able to speak. So the minimum our we're off this triangle issue goto a hamstring.