NIMCET 2012 Question Paper with solutions

The equation of an ellipse whose major axis along x-axis is
$\frac{{\mathit{x}}^2}{{\mathit{a}}^2}+\frac{{\mathit{y}}^2}{{\mathit{b}}^2}$ = 1 ... (i)
Eq, (i) passes through the points (4,3) and (-1,4) , then
$\frac{16}{{\mathit{a}}^2}+\frac{9}{{\mathit{b}}^2}$ = 1 ... (ii)
and $\frac{1}{{\mathit{a}}^2}+\frac{16}{{\mathit{b}}^2}$ = 1 ... (iii)
From Eqs. (ii) and (iii),
$16\left(1-\frac{16}{{\mathit{b}}^2}\right)$ + $\frac{9}{{\mathit{b}}^2}$ = 1
⇒ $\frac{9}{{\mathit{b}}^2}-\frac{256}{{\mathit{b}}^2}$ = = 1 - 16
⇒ $\frac{247}{{\mathit{b}}^2}$ = 15
∴ ${\mathit{b}}^2$ = $\frac{247}{15}$
From Eq. (iii)
$\frac{1}{{\mathit{a}}^2}$ = 1 - $\frac{16}{{\mathit{b}}^2}$ = 1 - $\frac{15}{247}$ × 16
⇒ $\frac{1}{{\mathit{a}}^2}$ = $\frac{247-240}{247}$ = $\frac{7}{247}$
⇒ $\left({\mathit{a}}^2=\frac{247}{7}\right)$
Now, put the value of ${\mathit{a}}^2$ and ${\mathit{b}}^2$ in Eq. (i) and get the required equation of an ellipse
$\frac{7{\mathit{x}}^2}{247}+\frac{15{\mathit{y}}^2}{247}$ = 1
⇒ $7{\mathit{x}}^2+15{\mathit{y}}^2$ = 247