Advertisement Remove all ads

# Find the Equation of an Ellipse Whose Axes Lie Along Coordinate Axes and Which Passes Through (4, 3) and (−1, 4). - Mathematics

Find the equation of an ellipse whose axes lie along coordinate axes and which passes through (4, 3) and (−1, 4).

Advertisement Remove all ads

#### Solution

$\text{ Let the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ and it passes through the points } (4,3) \text{ and } (-1,4).$
$\therefore \frac{16}{a^2}+\frac{9}{b^2} =1 \text{ and } \frac{1}{a^2}+\frac{16}{b^2} =1$
$\text{ Let } \alpha=\frac{1}{a^2}\text{ and } \beta=\frac{1}{b^2}$
$\text{ Then } 16\alpha + 9\beta = 1 \text{ and } \alpha + 16\beta = 1$
$\text{ Solving these two equations, we get }:$
$\alpha = \frac{7}{247} \text{ and } \beta = \frac{15}{247}$
$\Rightarrow \frac{1}{a^2} = \frac{7}{247}\text{ and } \frac{1}{b^2} = \frac{15}{247} . . . (1)$
$\text{Substituting eq. (1) in the equation of an ellipse, we get }:$
$\frac{7 x^2}{247} + \frac{15 y^2}{247} = 1$
$\text{ This is the required equation of the ellipse }.$

Is there an error in this question or solution?
Advertisement Remove all ads

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Exercise 26.1 | Q 14 | Page 23
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?