Question

Find the center, the lengths, the equations of the major and minor axes, the length and the equations of latus rectum, foci and directrices of the ellipse

25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0

.

Accepted Answer

Given ellipse $25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0$

$25 (x^{2} - 6 x) + 9 (y^{2} - 10 y) = - 225$

$25 (x^{2} - 6 x + 9) + 9 (y^{2} - 10 y + 25) = - 225 + 225 + 225$

$25 {(x - 3)}^{2} + 9 {(y - 5)}^{2} = 225$

$\frac{{(x - 3)}^{2}}{9} + \frac{{(y - 5)}^{2}}{25} = 1$ ......(i)

Shifting the origin to the point $(3, 5)$ we have $(x - 3) = X, (y - 5) = Y$ ....(ii)

Substituting for $x and y in (i)$ , we have $\frac{X^{2}}{9} + \frac{Y^{2}}{25} = 1$ ....(iii)

So, the centre of $\frac{X^{2}}{9} + \frac{Y^{2}}{25} = 1$ is $(0, 0) i . e X = 0, Y = 0$

Putting $X and Y$ in (ii), we get $x = 3, y = 5$

Therefore, centre of (i) is $(3, 5)$

Now, lengths of the semi-major and semi-minor axes are $5 and 3$ respectively

Therefore, lengths of major and minor axes are $10 and 6$

The equations of the major and minor axes of (iii) are $X = 0, Y = 0$ or using (ii), they are $x = 3, y = 5$

Length of the latus rectum are $= \frac{2 b^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5}$

Eccentricity $(e) = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$

The equation of latus recta of (iii) are $Y = \pm a e = \pm 5 \times \frac{4}{5} = \pm 4$

Using (ii), $y - 5 = \pm 4 i . e y = 9 and y = 1$

Coordinates of foci are $(0, \pm a e) = (0, \pm 5 \times \frac{4}{5}) = (0, \pm 4)$

So, $x - 3 = 0, y - 5 = \pm 9$ . Therefore coordinates are $(3, 9) and (3, 1)$

Now, equations of diretrices are $Y = \pm \frac{25}{4} i . e y - 5 = \pm \frac{25}{4}$

$y = 5 \pm \frac{25}{4}$

Therefore, equations are $4 y - 45 = 0 and 4 y + 5 = 0$

M L Aggarwal Solutions for Chapter: Conic Sections, Exercise 9: EXERCISE

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Conic Sections, Exercise 9: EXERCISE

Questions from M L Aggarwal Solutions for Chapter: Conic Sections, Exercise 9: EXERCISE with Hints & Solutions

Learn and Prepare for any exam you want !