Question

Find the equation to the ellipses, whose centres are the origin, whose axes are the axes of coordinates, and which pass through

(a) the points $(2,2)$ and $(3,1)$ and

(b) the points $(1,4)$ and $(-6,1)$.

Answer 1

The standard equation of an ellipse whose axes are the coordinates axes is
$\frac{(x-x_0{)}^2}{a^2}+\frac{(y-y_0{)}^2}{b^2}=1$ .....(I)
where 2a is the major axis length and 2b is the minor axis length.
In our question we have the origin as (0, 0) and axes as the coordinate axes so $x_0=0$ and $y_0=0$ in the above given equation. On substituting the values the equation of ellipse is as follows:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .....(II)
(a) Now we have a condition that the ellipse passes through the points (2, 2) and (3, 1) so these two points are the solutions of the above equation.
Substituting the point (2, 2) in the equation of ellipse we get,
$\Rightarrow \frac{2^2}{a^2}+\frac{2^2}{b^2}=1$
$\Rightarrow \frac{4}{a^2}+\frac{4}{b^2}=1$ .....(III)
Again substituting the point (3, 1) in the equation of an ellipse we get,
$\Rightarrow \frac{3^2}{a^2}+\frac{1^2}{b^2}=1$
$\Rightarrow \frac{9}{a^2}+\frac{1}{b^2}=1$ .....(IV)
Next we have to solve the equations (III) and (IV)
Multiplying (IV) by 4 we get the equation:
$\frac{36}{a^2}+\frac{4}{b^2}=4$ .....(V)
Subtracting (III) from (V) we get
$\Rightarrow \frac{36-4}{a^2}+\frac{4-4}{b^2}=4-1$
$\Rightarrow \frac{32}{a^2}=3$
$\Rightarrow a^2=\frac{32}{3}$
Substituting the value of $a^2$ in Equation (III) we get,
$\Rightarrow \frac{12}{32}+\frac{4}{b^2}=1$
$\Rightarrow \frac{4}{b^2}=1-\frac{12}{32}$
$\Rightarrow \frac{1}{b^2}=\frac{5}{32}$
$\Rightarrow b^2=\frac{32}{5}$
so the equation of the ellipse is :
$\frac{3x^2}{32}+\frac{5y^2}{32}=1$ .....Answer
(b)Now we have a condition that the ellipse passes through the points (1, 4) and (-6, 1) so these two points are the solutions of the above equation.
Substituting the point (1, 4) in the equation of ellipse we get,
$\Rightarrow \frac{1^2}{a^2}+\frac{4^2}{b^2}=1$
$\Rightarrow \frac{1}{a^2}+\frac{16}{b^2}=1$ .....(VI)
Again substituting the point (-6, 1) in the equation of an ellipse we get,
$\Rightarrow \frac{(-6{)}^2}{a^2}+\frac{1^2}{b^2}=1$
$\Rightarrow \frac{36}{a^2}+\frac{1}{b^2}=1$ .....(VII)
Multiplying equation (VII) with 16
$\frac{576}{a^2}+\frac{16}{b^2}=16$ .....(VIII)
Subtracting equation (VI) from (VIII) we get,
$\Rightarrow \frac{576-1}{a^2}+\frac{16-16}{b^2}=16-1$
$\Rightarrow \frac{575}{a^2}=15$
$\Rightarrow a^2=\frac{575}{15}$
Substituting the value of $a^2$ in Equation (I) we get,
$\Rightarrow \frac{15}{575}+\frac{16}{b^2}=1$
$\Rightarrow \frac{16}{b^2}=1-\frac{15}{575}$
$\Rightarrow \frac{1}{b^2}=\frac{35}{575}$
$\Rightarrow b^2=\frac{575}{35}$
Substituting $a^2$ and $b^2$ in the equation (II) we get,
$\frac{15x^2}{575}+\frac{35y^2}{575}=1$ .....Answer