Mathematics - Ellipse Question with Solution | TestHub

$5x^2+9y^2=45$ $\mathrm{a}=3,\mathrm{b}=\sqrt{5},\mathrm{e}=\frac{2}{3}$, one end of latus rectum in first quadrant $\left(2,\frac{5}{3}\right)$

Equation of tangent $2x+3y=9$

It meets axes at $\left(\frac{9}{2},0\right)$ and $(0,3)$

Area of the quadrilateral

$=4\cdot \frac{1}{2}\cdot \frac{9}{2}\cdot 3=27$

Let the equation of the ellipses ${\mathrm{E}}_2$ and ${\mathrm{E}}_1$ be

$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ and $\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{x}}^2}{{\mathrm{c}}^2}=1$

Figure (I) and figure (II) give the same area which is required.

Area of ${\mathrm{E}}_2=\pi \mathrm{ab}$ and Area of ${\mathrm{E}}_1=\pi \mathrm{bc}$

Required Area $=\pi \mathrm{b}(\mathrm{a}-\mathrm{c})$

Now, ${\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)$ and ${\mathrm{c}}^2={\mathrm{a}}^2{\left(1-{\mathrm{e}}^2\right)}^2$

$\Rightarrow \mathrm{c}=\mathrm{a}\left(1-{\mathrm{e}}^2\right)\Rightarrow \mathrm{a}-\mathrm{c}={\mathrm{ae}}^2$

Required Area $=\pi ab{e}^2$

$=9\times \frac{4}{9}=4$ sq. unit

(because $\pi \mathrm{ab}=\frac{1}{3}\times$ Area of quadratilateral)