Question

Find the equation of the directrix of the ellipse ${\displaystyle \frac{x^2}{100}}+{\displaystyle \frac{y^2}{36}}=1$ .

Answer 1

Hint: To find the equation of the directrix of the ellipse ${\displaystyle \frac{x^2}{100}}+{\displaystyle \frac{y^2}{36}}=1$ , we will compare this equation to the standard form ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ and find a and b. Then, we have to find the eccentricity using the formula $e=\sqrt{1-{\displaystyle \frac{b^2}{a^2}}}$ . We can find the directrix using the formula $x=\pm {\displaystyle \frac{a}{e}}$ . We have to substitute the values and form an equation.

Complete step by step answer:
We have to find the equation of the directrix of the ellipse ${\displaystyle \frac{x^2}{100}}+{\displaystyle \frac{y^2}{36}}=1$ . We know that the standard equation of an ellipse is given by ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ , where where a and b are the length of semi-major and semi-minor axis of an ellipse. Let us compare the given equation to this standard equation. We can see that $a^2=100$ and $b^2=36$ .
Let us consider $a^2=100$ . We have to take the square root of this equation.
$\begin{array}{rl}& \Rightarrow a=\sqrt{100}\\ & \Rightarrow a=10\text{ units}\end{array}$
Now, let us consider $b^2=36$ . We have to take the square root of this equation.
$\begin{array}{rl}& \Rightarrow b=\sqrt{36}\\ & \Rightarrow b=6\text{ units}\end{array}$
Therefore, we found a and b.
We know that eccentricity of an ellipse is given by
$e=\sqrt{1-{\displaystyle \frac{b^2}{a^2}}}$
Let us substitute the values of a and b in the above equation. We will get the eccentricity of the given ellipse as
$\Rightarrow e=\sqrt{1-{\displaystyle \frac{36}{100}}}$
Let us simplify the RHS.
$\begin{array}{rl}& \Rightarrow e=\sqrt{{\displaystyle \frac{100-36}{100}}}\\ & \Rightarrow e=\sqrt{{\displaystyle \frac{64}{100}}}\\ & \Rightarrow e={\displaystyle \frac{\sqrt{64}}{\sqrt{100}}}\end{array}$
We know that $\sqrt{64}=8$ and $\sqrt{100}=10$ . Therefore, we can write the above equation as
$\Rightarrow e={\displaystyle \frac{8}{10}}$
We have to cancel the common factor 2 from the RHS of the above equation.
$\Rightarrow e={\displaystyle \frac{4}{5}}$
We know that If an ellipse has centre $(0,0)$ ,eccentricity, e and semi-major axis, a in the x-direction, then the directrix is given by
$x=\pm {\displaystyle \frac{a}{e}}$
Let us substitute the value of a and e in the above equation.
$\begin{array}{rl}& \Rightarrow x=\pm {\displaystyle \frac{10}{{\displaystyle \frac{4}{5}}}}\\ & \Rightarrow x=\pm 10\times {\displaystyle \frac{5}{4}}\\ & \Rightarrow x=\pm {\displaystyle \frac{50}{4}}\end{array}$
Let us cancel the common factor of 2 from the RHS.
$\Rightarrow x=\pm {\displaystyle \frac{25}{2}}$
Let us take 2 from the RHS to the LHS to create an equation.
$\Rightarrow 2x=\pm 25$
Let us draw the graph of the given equation.

Hence, equation of the directrix of the ellipse ${\displaystyle \frac{x^2}{100}}+{\displaystyle \frac{y^2}{36}}=1$ is $2x=\pm 25$ .

Note: Students must be thorough with the formulas related to the ellipse. Here, we have used the standard form ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ , when a is along x-axis. We know that for an ellipse $a>b$ . The largest value (a) comes along with x in the given equation ${\displaystyle \frac{x^2}{100}}+{\displaystyle \frac{y^2}{36}}=1$ . If the equation was of the form ${\displaystyle \frac{x^2}{36}}+{\displaystyle \frac{y^2}{100}}=1$ , we would have used the standard equation ${\displaystyle \frac{x^2}{b^2}}+{\displaystyle \frac{y^2}{a^2}}=1$ and the corresponding directrix can be found using the formula $y=\pm {\displaystyle \frac{b}{e}}$ .