Question

Equation of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$

• A. $y=\pm \surd 5$
• B. $y=-\surd 5$
• C. $y=1\pm \surd 5$
• D. $x=-1\pm \surd 5$

Answer 1

The correct option is C

$y=1\pm \surd 5$

Step 1. Solve the given equation:

$9x^2+4y^2-18x-8y-23=0$

$\Rightarrow$${(3x-3)}^2+{(2y-2)}^2-13-23=0$

$\Rightarrow$ $9{(x-1)}^2+4{(y-1)}^2=36$

$\Rightarrow$ $\frac{{(x-1)}^2}{4}+\frac{{(y-1)}^2}{9}=1$

Step 2. Shifting origin to (1,1)

$\Rightarrow$ $x-1=X$

$y-1=Y$

$\Rightarrow$$\frac{X^2}{4}+\frac{Y^2}{9}=1$

Here, $a=2,b=3$

$\because$The eccentricity of the ellipse, ${\mathit{e}}^{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{}$

$=1-\frac{4}{9}$

$=\frac{5}{9}$

$\Rightarrow$ $e=\pm \sqrt{\frac{5}{9}}$

$=\pm \frac{\sqrt{5}}{3}$

$\because$The focus of the ellipse $=\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{\pm }\mathbf{}\mathit{b}\mathit{e}\mathbf{)}$

$=(0,\pm \sqrt{5})$

Step3: Find the latus rectum of the ellipse:

latus rectum will be ​$Y=\pm \sqrt{5}$

Shifting back, $Y=y-1$

$\Rightarrow$ $y-1=\pm \sqrt{5}$

$\Rightarrow$ $y=1\pm \sqrt{5}$

Thus, the equation of latus rectum is $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{5}}$

​ Hence, Option ‘C’ is Correct.