Question

If the end points of one axis of an ellipse are $(-12,4)$ and $(14,4)$ and eccentricity $\frac{12}{13}$, then the equation(s) of the ellipse is/are

• A. $\frac{(x-1{)}^2}{25}+\frac{(y-4{)}^2}{169}=1$
• B. $\frac{(x-1{)}^2}{169}+\frac{(y-4{)}^2}{25}=1$
• C. $(x-1{)}^2+\frac{25(y-4{)}^2}{169}=169$
• D. $(x-1{)}^2+\frac{(y-4{)}^2}{169}=169$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{(x-1{)}^2}{169}+\frac{(y-4{)}^2}{25}=1$
C $(x-1{)}^2+\frac{25(y-4{)}^2}{169}=169$

Centre is midpoint of vertices
$C\equiv (1,4)$
Now let the equation of the ellipse
$\frac{(x-1{)}^2}{a^2}+\frac{(y-4{)}^2}{b^2}=1$
Now putting the coordiantes of any vertex, we get
$\frac{{13}^2}{a^2}+0=1\Rightarrow a^2=169$
Now
Case $1:a>b$
Given $e=\frac{12}{13}$
$\Rightarrow e^2=1-\frac{b^2}{a^2}\Rightarrow b^2=25\therefore \text{ellipse will be}\frac{(x-1{)}^2}{169}+\frac{(y-4{)}^2}{25}=1$

Case $2:b>a$
Given $e=\frac{12}{13}$
$\Rightarrow e^2=1-\frac{a^2}{b^2}\Rightarrow b^2={\left(\frac{169}{5}\right)}^2\therefore \text{ellipse will be}(x-1{)}^2+\frac{25(y-4{)}^2}{169}=169$