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The point $(4,-3)$ with respect to the ellipse $4 x^2+5 y^2=1$
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Verified Answer
The correct answer is:
Is outside the curve
Using the condition the point $\left(x_1, y_1\right)$ lies
(i) On the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ if
$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1=0$
(ii) Outside the ellipse if $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\gt0$
(iii) Inside the ellipse if $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1 \lt 0$
Given ellipse is $\frac{x^2}{1 / 4}+\frac{y^2}{1 / 5}=1$
$\therefore \frac{16}{1 / 4}+\frac{9}{1 / 5}-1=64+45-1\gt0$
Point $(4,-3)$ lies outside the ellipse.
(i) On the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ if
$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1=0$
(ii) Outside the ellipse if $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\gt0$
(iii) Inside the ellipse if $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1 \lt 0$
Given ellipse is $\frac{x^2}{1 / 4}+\frac{y^2}{1 / 5}=1$
$\therefore \frac{16}{1 / 4}+\frac{9}{1 / 5}-1=64+45-1\gt0$
Point $(4,-3)$ lies outside the ellipse.
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