Question

The number of integral points which lies inside the ellipse whose vertices and foci are $(\pm 5,0)$ and $(\pm 4,0)$ respectively is

Answer 1

Vertices $(\pm a,0)\equiv (\pm 5,0)$
$\therefore a^2=25$
Foci
$(\pm ae,0)\equiv (\pm 4,0)\Rightarrow (ae{)}^2=16$
$\Rightarrow a^2-b^2=16$
$\therefore b^2=9$
$\therefore$ Equation of ellipse is
$\frac{x^2}{25}+\frac{y^2}{9}=1$

Now for the points which lie inside the ellipse and in first quadrant (other than axes) have to follow the inequality
$0<x<5,0<y<3$
hence possible points can be
$(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2)$
but from ellipse equation
For $x=3$
$\frac{9}{25}+\frac{y^2}{9}=1\Rightarrow |y|=\frac{12}{5}>2$
Hence $(3,2)$ is inside of ellipse
For $x=4$
$\frac{16}{25}+\frac{y^2}{9}=1\Rightarrow |y|=\frac{9}{5}<2$
Hence $(4,2)$ is not inside point of ellipse.
So points inside the ellipse in $1$st Quadrant will be
$(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1)$

$\therefore$ total number of points inside ellipse are
$=4\times 7+\text{points on axes}=28+[2\times 4+2\times 2+1]=41$