Question

By a suitable change of origin, show that the equation 9(x − 5)² + 16(y + 3)² = 144 represent an ellipse and that the coordinates of the centre are (5, −3). Find the position vectors of the foci and the ends of the major and minor axes.

Answer 1

$$9(x-5)^2+16(y+3)^2=144$$

$$\dfrac{(x-5)^2}{16}+\dfrac{(y+3)^2}{9}=1$$

The equation of ellipse

$$\dfrac{(x-a_C)^2}{a^2}+\dfrac{(y-y_C)^2}{b^2}=1$$

The center $C(5, -3)$

$$a=4, b=3$$

$$c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}=\sqrt{7}$$

Foci: $(5-\sqrt{7}, -3), (5+\sqrt{7}, -3).$

Ends of the major axis: $(1, -3), (9, -3).$

Ends of the minor axis: $(5, -6), (5, 0).$