Question

Find the coordinates of the centre, length of major and minor axes, equations of major and minor axes, eccentricity and length of latus rectum for the inclined ellipse $\frac{(5x-3y-9{)}^2}{20}+\frac{(3x+5y+9{)}^2}{32}=1$.

• A. Eccentricity $=\sqrt{\frac{3}{8}}$
• B. Length of latus rectum is $=\frac{5}{\sqrt{17}}$
• C. Equation of major and minor axes are $5x-3y-9=0$ and $3x+5y+9=0$ respectively.
• D. Length of major axis $=2\sqrt{\frac{16}{17}}$
  Length of minor axis $=2\sqrt{\frac{10}{17}}$
• E. Centre $\equiv (\frac{9}{17},-\frac{36}{17})$

Answer 1

Answer: (A), (B), (C), (D), (E)

Centre $\equiv (\frac{9}{17},-\frac{36}{17})$

Given: $\frac{(5x-3y-9{)}^2}{20}+\frac{(3x+5y+9{)}^2}{32}=1$

To Find:
$\left(i\right)$ Equations of major and minor axes
$\left(ii\right)$ Length of major and minor axes
$\left(iii\right)$ Eccentricity
$\left(iv\right)$ Length of latus rectum
$\left(v\right)$ Coordinates of the centre

Step - $1$: Recall the equation of the inclined ellipse in the $(X,Y)$ coordinate system.

Step - $2$: Simplify the given equation and find major and minor axes, equations of major and minor axes.

Step - $3$: Find the required parameters by comparison recall the formula of eccentricity, latus rectum.

Step - $4$: Find the common solution of equations of major and minor axes to get to the centre.

$\frac{(5x-3y-9{)}^2}{20}+\frac{(3x+5y+9{)}^2}{32}=1$ . . . $\left(1\right)$

We know that, the standard equation of an inclined ellipse is $\frac{(\frac{|a_{1}x+b_{1}y+c_1{|}^2}{\sqrt{a_1^2+b_1^2}})}{a^2}+\frac{(\frac{|b_{1}x-a_{1}y+c_2{|}^2}{\sqrt{a_1^2+b_1^2}})}{b^2}$

Rewriting $\left(1\right)$ in standard form.

Multiply and divide by $34$. $\because \sqrt{5^2+3^2}=\sqrt{34}$

$\frac{(5x-3y-9{)}^2}{20\times 34}\times 34+\frac{(3x+5y+9{)}^2}{32\times 34}\times 34=1$

$\left(i\right)$ Equations of major and minor axes

$\Rightarrow \frac{(\frac{|5x-3y-9{|}^2}{\sqrt{34}})}{\frac{20}{34}}+\frac{(\frac{|3x+5y+9{|}^2}{\sqrt{34}})}{\frac{32}{34}}$

$\Rightarrow \frac{(\frac{|5x-3y-9{|}^2}{\sqrt{34}})}{\frac{10}{17}}+\frac{(\frac{|3x+5y+9{|}^2}{\sqrt{34}})}{\frac{16}{17}}$

$\Rightarrow a^2=\frac{16}{17}$ and $b^2=\frac{10}{17}$

$\Rightarrow X=\frac{|3x+5y+9|}{\sqrt{34}}$, $Y=\frac{|5x-3y-9|}{\sqrt{34}}$

$\Rightarrow a=\sqrt{\frac{16}{17}}$ and $b=\sqrt{\frac{10}{17}}$, $a>b$

Thus, $Y=0$ is the major axis and $X=0$ is the minor axis.

$5x-3y-9=0$ equation of major axis.

$3x+5y+9=0$ equation of minor axis.

$\left(ii\right)$ Length of major and minor axes

$\Rightarrow a=\sqrt{\frac{16}{17}}$ and $b=\sqrt{\frac{10}{17}}$, $a>b$

Length of major axis $=2a=2\sqrt{\frac{16}{17}}$

Length of minor axis $=2b=2\sqrt{\frac{10}{17}}$

$\left(iii\right)$ Eccentricity

$e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\frac{\frac{10}{17}}{\frac{16}{17}}}$

$\Rightarrow e=\sqrt{1-\frac{10}{16}}=\sqrt{\frac{3}{8}}$

$\left(iv\right)$ Length of latus rectum

$\frac{2b^2}{a}=\frac{2\times \frac{10}{17}}{\sqrt{\frac{16}{17}}}=\frac{20}{4\sqrt{17}}$

$=\frac{5}{\sqrt{17}}$

$\left(v\right)$ Coordinates of the centre

$5x-3y-9=0$ equation of major axis.

$3x+5y+9=0$ equation of minor axis.

Find their intersection.

$x=\frac{9}{17}$ and $y=-\frac{36}{17}$

Centre $\equiv (\frac{9}{17},-\frac{36}{17})$