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10-2. Parabola, Ellipse, Hyperbola
easy
The point $(4, -3)$ with respect to the ellipse $4{x^2} + 5{y^2} = 1$
A
Lies on the curve
B
Is inside the curve
C
Is outside the curve
D
Is focus of the curve
Solution
(c) Using the condition the point $({x_1},\,{y_1})$ 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 $\frac{{x_1^2}}{{{a^2}}} + \frac{{y_1^2}}{{{b^2}}} – 1 > 0$
$(iii)$ Inside the ellipse if $\frac{{x_1^2}}{{{a^2}}} + \frac{{y_1^2}}{{{b^2}}} – 1 < 0$
Given ellipse is $\frac{{{x^2}}}{{1/4}} + \frac{{{y^2}}}{{1/5}} = 1$
$\frac{{16}}{{1/4}} + \frac{9}{{1/5}} – 1 = 64 + 45 – 1 > 0$
Point $(4, -3)$ lies outside the ellipse.
Standard 11
Mathematics
