Position of point (1, 3) with respect to ellipse 4{x^2} + 9{y^2} - 16x - 54y + 61 = 0

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10-2. Parabola, Ellipse, Hyperbola

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The position of the point $(1, 3)$ with respect to the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$

A

Outside the ellipse

B

On the ellipse

C

On the major axis

D

On the minor axis

Solution

(c) $E \equiv 4 + 9{(3)^2} – 16(1) – 54(3) + 61 < 0$

Therefore, the point is inside the ellipse.

$\frac{{4{{(x – 2)}^2}}}{{36}} + \frac{{9{{(y – 3)}^2}}}{{36}} = 1$

Equation of major axis is $y – 3 = 0$ and point $(1, 3)$ lies on it.

Standard 11

Mathematics

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