Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The distance L between the centre and any normal in the ellipse is Option: 1 <img alt="\large L>a+b" src="https://learn.careers360.com/latex-

The distance L between the centre and any normal in the ellipse is 

Option: 1

\large L>a+b


Option: 2

\large L>a-b


Option: 3

\large L\in(a-b,a+b)


Option: 4

\large L <(a-b)


Answers (1)

best_answer

The equation of any normal to the ellipse at P(\theta) is

a x \sec \theta-b y \csc \theta=a^{2}-b^{2}

Let C be the center of the ellipse

Then CM = Length of the perpendicular from the center C to the normal

\large =\frac{\left(a^{2}-b^{2}\right)}{\sqrt{a^{2} \sec ^{2} \theta+b^{2} \csc ^{2} \theta}}

\large =\frac{\left(a^{2}-b^{2}\right)}{\sqrt{a^{2}+b^{2}+a^{2} \tan ^{2} \theta+b^{2} \cot ^{2} \theta}}

\large <\frac{\left(a^{2}-b^{2}\right)}{\sqrt{a^{2}+b^{2}+2 a b}}=a-b

 

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions