Question

How do you find the equation of the ellipse that passes through the points $\left( {6,4} \right)$ and $\left( { - 8,3} \right)$ ?

Answer 1

In the given question, we are required to find the equation of an ellipse passing through two given points $\left( {6,4} \right)$ and $\left( { - 8,3} \right)$. We are not given the center of the ellipse in the problem, so we will consider it to be the origin. Then, we will substitute the two points given to us as they satisfy the equation of ellipse.

Complete step by step solution:
So, we shall take the center of the ellipse as origin.
The equation of the ellipse $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$ where $\left( {h,k} \right)$ is the centre of the ellipse.
So, substituting the center of the ellipse, we get,
$\dfrac{{{{\left( {x - 0} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - 0} \right)}^2}}}{{{b^2}}} = 1$
$\Rightarrow \dfrac{{{{\left( x \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( y \right)}^2}}}{{{b^2}}} = 1$
The equation above is the equation of an ellipse having origin as center.
The given ellipse passes through points $\left( {6,4} \right)$ and $\left( { - 8,3} \right)$.
Putting $\left( {6,4} \right)$ in the equation $\dfrac{{{{\left( x \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( y \right)}^2}}}{{{b^2}}} = 1$, we get,
$\Rightarrow \dfrac{{{{\left( 6 \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( 4 \right)}^2}}}{{{b^2}}} = 1$
$\Rightarrow \dfrac{{36}}{{{a^2}}} + \dfrac{{16}}{{{b^2}}} = 1 - - - - - - \left( 1 \right)$
Putting $\left( { - 8,3} \right)$ in the equation $\dfrac{{{{\left( x \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( y \right)}^2}}}{{{b^2}}} = 1$, we get,
$\Rightarrow \dfrac{{{{\left( { - 8} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( 3 \right)}^2}}}{{{b^2}}} = 1$
$\Rightarrow \dfrac{{64}}{{{a^2}}} + \dfrac{9}{{{b^2}}} = 1 - - - - - - \left( 2 \right)$
Multiplying equation $\left( 1 \right)$ with $9$ and equation $\left( 2 \right)$ with $16$, we get,
$9\left( {\dfrac{{36}}{{{a^2}}} + \dfrac{{16}}{{{b^2}}}} \right) = 9$ and $16\left( {\dfrac{{64}}{{{a^2}}} + \dfrac{9}{{{b^2}}}} \right) = 16$
$\Rightarrow \dfrac{{324}}{{{a^2}}} + \dfrac{{144}}{{{b^2}}} = 9 - - - - \left( 3 \right)$ and $\dfrac{{1024}}{{{a^2}}} + \dfrac{{144}}{{{b^2}}} = 16 - - - - - \left( 4 \right)$
Subtracting equation $\left( 3 \right)$ from equation $\left( 4 \right)$, we get,
$\dfrac{{1024}}{{{a^2}}} + \dfrac{{144}}{{{b^2}}} - \dfrac{{324}}{{{a^2}}} - \dfrac{{144}}{{{b^2}}} = 16 - 9$
Cancelling the like terms with opposite signs, we get,
$\Rightarrow \dfrac{{700}}{{{a^2}}} = 7$
Taking ${a^2}$ to right side of the equation and dividing both sides of the equation by $7$, we get,
$\Rightarrow {a^2} = 100$
Taking square root of both sides of the equation, we get,
$\Rightarrow a = 10$
We neglect $a = - 10$ as a represents the length of the major axis of the ellipse and hence, cannot be negative.
Now, putting the value of a in equation $\left( 1 \right)$, we get,
$\Rightarrow \dfrac{{36}}{{{{10}^2}}} + \dfrac{{16}}{{{b^2}}} = 1$
$\Rightarrow \dfrac{{36}}{{100}} + \dfrac{{16}}{{{b^2}}} = 1$
Simplifying further, we get,
$\Rightarrow \dfrac{{16}}{{{b^2}}} = 1 - \dfrac{{36}}{{100}}$
$\Rightarrow \dfrac{{16}}{{{b^2}}} = \dfrac{{64}}{{100}}$
Isolating b so as to find b,
$\Rightarrow {b^2} = 25$
Taking square root of both sides of the equation, we get,
$\Rightarrow b = 5$
We neglect $b = 5$ as a represents the length of the minor axis of the ellipse and hence, cannot be negative.

So, the equation of the ellipse $\dfrac{{{x^2}}}{{100}} + \dfrac{{{y^2}}}{{25}} = 1$

Note: There are various methods of finding equations of ellipse when certain parameters are given to us. Different conditions require us to solve the problem through different methods. It is not necessary that all the ellipse would either be along the x axis or y axis. Care should be taken while handling problems involving oblique ellipse.