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