3 Marks Question

Question 13 Marks

Find the equation of the ellipse in the follwing case:focus is (-2, 3), directrix is 2x + 3y + 4 = 0 and $\text{e}=\frac{4}{5}.$

Answer

Let P (x, y) be a point the ellipse. Then, by definition SP = e PM Here $\text{e}=\frac{4}{5},$ coordinates of S are (-2, 3) and the equation of directrix is 2x + 3y + 4 = 0$\therefore\text{ SP}=\frac{4}{5}\text{PM}$
$\Rightarrow\text{SP}^2=\frac{16}{25}(\text{PM})^2$
$\Rightarrow25\text{ SP}^2=16\text{ PM}^2$
$\Rightarrow25\bigg[\big(\text{x}+2\big)^2+\big(\text{y}-3\big)^2\bigg]=16\bigg[\frac{2\text{x}+3\text{y}+4}{\sqrt{2^2+3^2}}\bigg]^2$
$\Rightarrow25\big[\text{x}^2+4+4\text{x}+\text{y}^2+9-6\text{y}\big]=\frac{16\big(2\text{x}+3\text{y}+4\big)^2}{13}$
$\Rightarrow325\big[\text{x}^2+\text{y}^2+4\text{x}-6\text{y}+13\big]=16\big(2\text{x}+3\text{y}+4\big)^2$
This is the required equation of the ellipse.

View full question & answer→

Question 23 Marks

Find the equation to the ellipse whose centre is $(-2, 3)$ and semi-axis are $3$ and $2$ when major axis is Parallel to $y-$axis.

Answer

Let $2a$ and $2b$ the major and minor axes of the ellipse.
Then, its equation is$\frac{(\text{x}+2)^2}{\text{a}^2}+\frac{(\text{y}-3)^2}{\text{b}^2}=1$
$\big[\because\ $centre$: (-2, 3) \dots(\text{i})\ \big]$
we have, semi-major axis $= a = 2\Rightarrow\text{a}^2=4$
and semi-major axis $= b = 3\Rightarrow\text{b}^2=4$
Putting $a^2 = 4$ and $b^2 = 9$ in equation $(i),$ we get$\frac{(\text{x}+2)^2}{4}+\frac{(\text{y}-3)^2}{9}=1$
$\Rightarrow\frac{9(\text{x}+2)^2+4(\text{y}-3)^2}{36}=1$
$\Rightarrow9(\text{x}+2)^2+4(\text{y}-3)^2=36$
$\Rightarrow9\big[\text{x}^2+4+4\text{x}\big]+4\big[\text{y}^2+9-6\text{y}\big]=36$
$\Rightarrow9\text{x}^2+36+36\text{x}+4\text{y}^2+36-24\text{y}=36$
$\Rightarrow9\text{x}^2+4\text{y}^2+36\text{x}-24\text{y}+36+36-36=0$
$\Rightarrow9\text{x}^2+4\text{y}^2+36\text{x}-24\text{y}+36=0$

View full question & answer→

Question 33 Marks

Find the eccentricity of an ellipse whose latus-rectum ishalf of its Major axis.

Answer

Let 2a and 2b be the major axes of the ellipse, when latus-rectum is half of major-axis.$\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{b}$
$\Rightarrow2\text{b}^2=\text{a}^2$
$\Rightarrow\text{a}^2=2\text{b}^2$
Now, $\text{b}^2=\text{a}^2\big(1-\text{e}^2\big)$
$\Rightarrow{\text{b}^2}=2\text{b}^2\big(1-\text{e}^2\big)$ $\big[\because\text{a}^2=2{\text{b}^2}\big]$
$\Rightarrow1=2\big(1-\text{e}^2\big)$
$\Rightarrow1=2-2\text{e}^2$
$\Rightarrow2\text{e}^2=2-1$
$\Rightarrow\text{e}^2=\frac{1}{2}$
$\Rightarrow\text{e}=\frac{1}{\sqrt{2}}$

View full question & answer→

Question 43 Marks

Find the eccentricity of an ellipse whose latus-rectum ishalf of its minor axis

Answer

Let 2a and 2b be the major axes of the ellipse, when latus-rectum is half of minor axis.$\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{b}$
$\Rightarrow2\text{b}^2=\text{ab}$
$\Rightarrow\frac{\text{b}^2}{b}=\frac{\text{a}}{2}$
$\Rightarrow\text{b}=\frac{\text{a}}{2}$
$\Rightarrow\text{b}^2=\frac{\text{a}^2}{4}$
Now, $\text{b}^2=\text{a}^2\big(1-\text{e}^2\big)$
$\Rightarrow\frac{\text{a}^2}{4}=\text{a}^2\big(1-\text{e}^2\big)$ $\Big[\because\text{b}^2=\frac{\text{a}^2}{4}\Big]$

View full question & answer→

Maths 3 Marks Question

Take a timed test