Question

The equations of the latus rectum of the ellipse $9{x^2} + 4{y^2} - 18x - 8y - 23 = 0$ are
A) $y = \pm \sqrt 5$
B) $y = - \sqrt 5$
C) $y = 1 \pm \sqrt 5$
D) $y = - 1 \pm \sqrt 5$

Answer 1

We will first find the standard equation of the ellipse by completing the squares in the given equation. Then we will shift the origin and then we will find the eccentricity of the ellipse using the respective formula. Finally, using the eccentricity we will find the equations of the latus rectum.

Formula used:
Eccentricity, ${e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}}$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes respectively.

Complete step by step solution:
The given equation of the ellipse is $9{x^2} + 4{y^2} - 18x - 8y - 23 = 0$.
Let us bring this equation into the standard form of the equation of an ellipse. For this, we will collect the like terms together. We get
$(9{x^2} - 18x) + (4{y^2} - 8y) - 23 = 0$ ………$(1)$
Now, we have to complete the squares on the LHS.
Let us first complete the square of $9{x^2} - 18x$.
We know that $9{x^2} - 18x = {(3x)^2} - 2 \times 3x \times 3$. So, to complete the square, we have to add and subtract ${3^2}$. This gives us
$9{x^2} - 18x + {3^2} - {3^2} = 9({x^2} - 2x + 1) - 9$
$\Rightarrow 9{x^2} - 18x = 9{(x - 1)^2} - 9$ ……..$(2)$
Similarly, when we complete the square of $4{y^2} - 8y$, we get
$4{y^2} - 8y = 4{(y - 1)^2} - 4$ ………$(3)$
Substituting equations $(2)$ and $(3)$ in equation $(1)$, we have
$\begin{array}{l}9{(x - 1)^2} - 9 + 4{(y - 1)^2} - 4 - 23 = 0\\\ \Rightarrow 9{(x - 1)^2} + 4{(y - 1)^2} = 36\end{array}$
Dividing both sides of the above equation by 36, we get the equation of ellipse as
$\Rightarrow \dfrac{{{{(x - 1)}^2}}}{4} + \dfrac{{{{(y - 1)}^2}}}{9} = 1$ ………$(4)$
The major axis of the ellipse is along the $y -$ axis and the minor axis is along the $x -$ axis. Comparing the above equation to the general equation of an ellipse $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$, we have ${a^2} = 9$ and ${b^2} = 4$. This means that $a = 3$ and $b = 2$.
In equation $(4)$, the origin of the ellipse is $(1,1)$.
Let us shift the origin to $(0,0)$ by using $X = x - 1$ and $Y = y - 1$ in equation $(4)$. Thus, equation $(4)$ becomes
$\dfrac{{{X^2}}}{4} + \dfrac{{{Y^2}}}{9} = 1$
Let us now find the eccentricity. We know that ${e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}}$. Substituting the values ${a^2} = 9$ and ${b^2} = 4$, we get
${e^2} = 1 - \dfrac{4}{9} = \dfrac{5}{9}$
$\Rightarrow e = \pm \sqrt {\dfrac{5}{9}}$
Since the eccentricity is never a negative value, we have $e = \dfrac{{\sqrt 5 }}{3}$
Hence, the foci of the given ellipse are ${F_1}(0,ae)$ and ${F_2}(0, - ae)$. Here, $ae = 3 \times \dfrac{{\sqrt 5 }}{3} = \sqrt 5$.
Thus, the foci are ${F_1}(0,\sqrt 5 )$ and ${F_2}(0, - \sqrt 5 )$.
The latus rectum passes through the foci. So, the equations of the latus rectum of the given ellipse are $Y = \sqrt 5$ and $Y = - \sqrt 5$ i.e., $Y = \pm \sqrt 5$.
Shifting the origin back to $(1,1)$, we have
$\begin{array}{l}y - 1 = \pm \sqrt 5 \\\ \Rightarrow y = 1 \pm \sqrt 5 \end{array}$

Therefore, the correct option is C.

Note:
When the equation of the ellipse is not in the standard form, we must use the method of completing the squares to bring it back to the standard form. We need to take care while identifying the major axis and the minor axis. Also, we need to keep in mind that the eccentricity of an ellipse is always a value which is less than 1, but always positive.