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
106 views
in Parabola by (240k points)
closed by
Determine the equation for the ellipse if center is at (0, 0), the major axis on the y-axis and ellipse passes through the points (4, 3) and (2, 5)
1. \(\rm 4y^2 + 3x^2 = 1\)
2. \(\rm 3y^2 + 4x^2 = 1\)
3. \(\rm 4y^2 + 3x^2 = 91\)
4. \(\rm 3y^2 + 4x^2 = 91\)

1 Answer

0 votes
by (238k points)
selected by
 
Best answer
Correct Answer - Option 4 : \(\rm 3y^2 + 4x^2 = 91\)

Concept:

The standard equation of an ellipse:

\(\rm {x^2\over a^2}+{y^2\over b^2} = 1\)

Where 2a and 2b are the length of the major axis and minor axis respectively and center (0, 0)

Calculation:

Given: Center is (0, 0)

So standard equation \(\rm {x^2\over a^2}+{y^2\over b^2} = 1\)

Since (4, 3) lies on ellipse satisfy the equation of ellipse

 \(\rm {16\over a^2}+{9\over b^2} = 1\)                  ...(i)

Similarly for (2, 5)

\(\rm {4\over a^2}+{25\over b^2} = 1\)                      ...(ii)

Substracting the equations (i) from 4 × (ii)

\(\rm {4\times 25\over b^2}-{9\over b^2} = 4-1\)

⇒ \(\rm {91\over b^2} = 3\)

⇒ \(\boldsymbol{\rm b^2 = {91\over3}}\)

Putting it back in equation (i)

\(\rm {16\over a^2}+{9\times 3\over 91} = 1\)

⇒ \(\rm {16\over a^2} =1-{27\over 91}\)

⇒ \(\rm {16\over a^2} = {64\over 91}\)

⇒ \(\boldsymbol{\rm a^2 = {91\over4}}\)

Equation of ellipse 

\(\rm {4x^2\over 91}+{3y^2\over 91} = 1\)

⇒ 4x2 + 3y2 = 91

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

...