The point $(1,3)$ with respect to the ellipse $4 x^2+9 y^2-16 x-54 y+61=0$ lies
The point $(1,3)$ with respect to the ellipse $4 x^2+9 y^2-16 x-54 y+61=0$ lies
outside the ellipse
on the ellipse
on the minor axis
on the major axis
Solution
Given, equation of ellipse is
$
4 x^2+9 y^2-16 x-54 y+61=0
$
It can be written as,
$
\frac{(x-2)^2}{9}+\frac{(y-3)^2}{4}=1
$
Centre of ellipse $(2,3)$.
vertex are $(5,3),(-1,3)$
Focus are $(2+\sqrt{5}, 3),(2-\sqrt{5}, 3)$
$\therefore(1,3)$ lies on major axis
