Question

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Let $f(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1$. To determine whether the point $(x_1,y_1)$ lies inside the ellipse, the necessary condition is:

• A. $f(x_1,y_1)<0$
• B. $f(x_1,y_1)>0$
• C. $f(x_1,y_1)=0$
• D. None of these

Answer 1

The correct option is A $f(x_1,y_1)<0$

$f(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1$ ........ $(1)$

The region (disk) bounded by the ellipse is given by the equation:

$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}\le 1$ centered at $(h,k)$.

The given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ centered at origin i.e. $(0,0)$.

The region bounded by this ellipse is

$\frac{(x{)}^2}{a^2}+\frac{(y{)}^2}{b^2}\le 1$ ...... $\left(1\right)$

The point $x_1,y_1$ lies inside the given ellipse if it satisfies $\left(1\right)$

i.e. if $\frac{(x_1{)}^2}{a^2}+\frac{(y_1{)}^2}{b^2}\le 1$ ...... $\left(1\right)$

if $\frac{(x_1{)}^2}{a^2}+\frac{(y_1{)}^2}{b^2}-1<0$ ...... $\left(1\right)$

$⟹$ $f(x_1,y_1)<0$ ....... From $\left(1\right)$

Hence, option A is correct.