Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
631 views
in Coordinate Geometry by (43.0k points)
closed by

Find the equation of the ellipse in standard form if: eccentricity is 2/3 and passes through (2, -5/3)

1 Answer

+1 vote
by (42.5k points)
selected by
 
Best answer

  Let the required equation of ellipse be \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) where a > b.

Given, eccentricity (e) = 2/3

The ellipse passes through (2, -5/3).

Substituting x = 2 and y = -5/3 in equation of ellipse, we get

The required equation of ellipse is \(\frac {x^2}{9} + \frac {y^2}{5} =1\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...