consider-the-tangent-line-to-the-ellipse-dfracx2a2-dfracy2b2-1-at-a-point-p-q-in-the-first-quadrant-

Question

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 $.

Wen Zheng · Accepted Answer

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Score since P squared plus C squared over Esquire, ask your square or be square is a good one. So we have you, Squire? Yes, he called you. You square for a square. My client a square minus p squared on behalf our squire is before too A to the park off or overpay squire Plus the square A square over a squared minus the square Here we denote X is equal to p squared Then take a derivative off square about axe. No I&#x27;LL square X This is control here I just write down the without esquire over a squire, minus x square hands. Exploiter comes you square minus a square. I&#x27;m slacks, Squire us to a To the public for thanks minus into the power of six. It&#x27;s the call to zero. So if we liked the narrative is equal to zero. We have p squire minus stay Square Ham&#x27;s exploiter us two to four acts minus a two six. This leak onto a serial, not his. That axe is ego too. He squares. This is greeted and cereal and smaller than a square self. This equation we have Max, this is his squire. Is they come true? Is cube over B plus? No, on. Actually, this&#x27;s culture the money they have Zac&#x27;s us A cube terms plus a sax minus is cube. This part is always graters. And zero since accident is between cereal and a square, Then we&#x27;re half If axe its greatest on this number the derivative It&#x27;s true that zero if it is smaller than this number directive is less than zero. So what actually is a call to this number? Our square has some minimal my value in this case we have How squire is the coach you a war over peace. Choirs of P squared is equal to you Face cube over the past day Anythin plus the squire square over Esquire minus is cube over B plus, eh? This&#x27;s equal. Chu, you&#x27;re just right down there without a plus B squared. We have hour minimum is equal to a class B. So part of sea is finding It&#x27;s a minimal area after the triangle. So Isis, we used eight Didn&#x27;t notice the area of the triangle This triangle Ariel to strangle We have a go too a half times the square over he comes he squire over Hugh in orderto find the minimal value off, eh? What we need to do is find the maximum value off, Pete, have ask you since A and B A constant number. Oh, yeah, I write Pete, I&#x27;m secu is equal to f This is a culture here we use Half square is equal to the square Q Squire P Square Timescale square is gone, too. You he squared hams E square over a square, a square minus p squared. This is equal to the square over a square. Pam&#x27;s he&#x27;s square square manis pedis to power off or derogative of, uh, square we have. Is he going to be square over Esquire hams to a square dance piece minus four times Pearce killed. If we like it, this narrative is equal to zero. We have, Yeah, I caught you a over squadron. Tough, too, and we can see one. He is greeted on this number. Half square. Prom is that&#x27;s zero. 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&#x27;s a maximum value, a square. My small value is equal to the squire over a square hams. It&#x27;s four over two minus to four Oh, for this is because you&#x27;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&#x27;re off this triangle issue goto a hamstring.

Problem

The frame for the kite is to be made from six pie…

Problem 67 Hard Difficulty

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

Calculus: Early Transcendentals

Anna Marie V.

Heather Z.

Kayleah T.

Michael J.

Integration Review - Intro

Definite vs Indefinite

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!

Calculus: Early Transcendentals

Anna Marie V.

Heather Z.

Kayleah T.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Anna Marie V.

Heather Z.

Kayleah T.

Michael J.

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!