Problem

Find the value of the number $ a $ such that the …

02:08

Problem 69 Hard Difficulty

Show that the ellipse $ x^2/a^2 + y^2/b^2 = 1 $ and the hyperbola $ x^2/A^2 - y^2/B^2 = 1 $ are orthogonal trajectories if $ A^2 < a^2 $ and $ a^2 - b^2 = A^2 + B^2 $ (so the ellipse and hyperbola have the same foci)

Name: show that the ellipse x2a2 y2b2 1 and the hyperbola x2a2 y2b2 1 are orthogonal trajectories if a2 a2
Uploaded: 2021-08-26
Channel: Numerade
Description: Show that the ellipse $ x^2/a^2 + y^2/b^2 = 1 $ and the hyperbola $ x^2/A^2 - y^2/B^2 = 1 $ are orthogonal trajectories if $ A^2 < a^2 $ and $ a^2 - b^2 = A^2 + B^2 $ (so the ellipse and hyperbola have the same foci)

Answer

ellipse and hyperbola are orthogonal trajectories.

November 18, 2017

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Askjfas S.

November 18, 2017

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Askjfas S.

November 18, 2017

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Video Transcript

we have a question we need to show that access choir by a Squire plus Y squared but being square equal to one. And the hyper bola. This is the lips and hyper bola access squired by a square minus y squared but be square. Well to one article. Electronic trees are total project remain means if a find slope of this divide by dx and slope of this let us right slope of tended on this which is uh if her lips and D Y by dx slope of hyper Ebola so productive they should be minus one. This is the meaning of all to the military secretary over here. So let us find dy dx differentiate with respect to X two x by a square plus two. Y. Do I buy DX Bible Square Equal to 0? Which means divided by D X will be equal to minus two X by a Squire into being squired by to Y. Two and two will get cancelled out. So do you have any excess hm minus X. B squired by esquire white. This is do you have any exit E. For ellipse? Similarly for help. Ebola. If we differentiate will be getting two X by a square -2 Whereby Be a Squire. Diva Body x equal to zero. Okay so uh D Y by D X will be equal to two x by a Squire into b squared by do it. These two will get cancelled out. So be esquire X by uh this is a Squire. Why this is D Y by the X slope of tangent at hyper bola. If we do like this we'd be getting uh minus X. Being squired by is quite why multiplied by be a square X by is quite a way. They should be equal to minus of one. Okay. Which means this is quite a way which means uh B squared B squared minus X squared equal to minus is square is square. Well y squared some mines mines will get cancelled out. So we have is it be as quiet right divided by a square is quite equal to Why? Square by Expert Equation # one. Now from any question let us get the value of access required by Y Squire. If he uh could just get the values. Mhm Okay, is everything good? Yes. Yeah. Let us get values from any one of these. So let us uh to get the values, we will be subtracting if this is a question number one, the same question number two. For the sake of simplicity will subtract any question number one from the question number two. The art form will be doing that technique. Question # two from the question # one. So this will be access choir by a square plus. We're squired by being square equal to one and minus. Access square by square. Mine's white square but it is quite equal to one minus plus and minus. So one will become right inside it will become zero and access square will be taken as common. So one by a square -1. But capital is squared bless. Why square will be taken as common? One by B squared minus one by capital B square equal to zero. Yeah. Uh so this is extra square and a square minuses square by square is what equal to Y squared one by b squared minus one by small, B squared Y squared more B squared minus capital B square being square. So we need to find the value of this. X squared, Y Y squared by X squared from here was squired by esquire will be able to is squared squared. Okay, we are dividing both sides by access. Quite so yeah, and this will be equal to be a squared B squared by a square square into is Squire minus a squared by small B squared minus capital base question Okay now we'll be plugging in the value in equation number let us say question number three in question number three. So we'll be getting could be a squared B squared by a square. A square equal to this being squared, B squared by Smalley square. Capital is squared square minus a squared by the square minus business. So these people get canceled out Here. It will be one. So the cross multiplying will be getting capital a square mind. Smalley square equal to a small base square minds Capital base square. Okay, so if we add capital D squared to both the sides, A squared plus b squared and again smaller square to both the sides. So this will become a squared minus B squared. So this is bro, thank you.

Suman Saurav T.

Chandigarh University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Problem

Find the value of the number $ a $ such that the …

Problem 69 Hard Difficulty

Show that the ellipse $ x^2/a^2 + y^2/b^2 = 1 $ and the hyperbola $ x^2/A^2 - y^2/B^2 = 1 $ are orthogonal trajectories if $ A^2 < a^2 $ and $ a^2 - b^2 = A^2 + B^2 $ (so the ellipse and hyperbola have the same foci)

Answer

ellipse and hyperbola are orthogonal trajectories.

More Answers

Related Courses

Calculus: Early Transcendentals

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