Question

Determine position of a point $(2,3)$ with respect to the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$.

• A. Outside
• B. Inside
• C. On the ellipse
• D. None of the above

Answer 1

The correct option is B Inside

Given equation is $\frac{x^2}{16}+\frac{y^2}{25}=1$

$\frac{x^2}{16}+\frac{y^2}{25}-1=0$

The curve is defined by,

$f(x,y)=\frac{x^2}{16}+\frac{y^2}{25}-1f(2,3)=\frac{{\left(2\right)}^2}{16}+\frac{{\left(3\right)}^2}{25}-1f(2,3)=\frac{1}{4}+\frac{9}{25}-1=\frac{25+36-100}{100}=-\frac{39}{100}f(2,3)<0$

So, the point lies inside the ellipse.

Option B is correct.