MATHS 5 Marks Questions

This is of the form $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where $\text{a}^2=\frac{1}{4}$ and $\text{b}^2=\frac{1}{9},\text{ i}.\text{e}.\text{a}=\frac{1}{2}$and $\text{b}=\frac{1}{3}.$

Clearly a > b

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

$\Rightarrow\text{e}=\sqrt{1-\frac{\frac{1}{9}}{\frac{1}{4}}}$

$\Rightarrow\text{e}=\sqrt{1-\frac{4}{9}}$

$\Rightarrow\text{e}=\frac{\sqrt{5}}{3}$

Coordinates of the foci $=(\pm\text{ae, 0})=\Big(\pm\frac{\sqrt{5}}{6},0\Big)$

Length of the latus rectum $=\frac{2\text{b}^2}{\text{a}}$

$=\frac{2\times\frac{1}{9}}{\frac{1}{2}}$

$=\frac{4}{9}$

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

Since the ellipse passes through

(1, 4) and (-6, 1).

$\therefore\ \frac{(1)^2}{\text{a}^2}+\frac{(4)^2}{\text{b}^2}=1$

$\Rightarrow\frac{1}{\text{a}^2}+\frac{16}{\text{b}^2}=1$

$\Rightarrow\text{b}^2+16\text{a}^2=\text{a}^2\text{b}^2\ \dots(\text{ii})$

and $\frac{(-6)^2}{\text{a}^2}+\frac{(1)^2}{\text{b}^2}=1$

$\Rightarrow\frac{36}{\text{a}^2}+\frac{1}{\text{b}^2}=1$

$\Rightarrow\frac{36}{\text{a}^2}+\frac{1}{\text{b}^2}=1$

$\Rightarrow36\text{b}^2+\text{a}^2=\text{a}^2\text{b}^2\ \dots\text{(iii)}$

Multipliying equation (iii) by 16, we get

$576\text{b}^2+16\text{a}^2=16\text{a}^2\text{b}^2\ \dots(\text{iv})$

Substituting equation (ii) from eqation (iv), we get

$576\text{b}^2-\text{b}^2=16\text{a}^2\text{b}^2-\text{a}^2\text{b}^2$

$\Rightarrow575\text{b}^2=15\text{a}^2\text{b}^2$

$\Rightarrow575=15\text{a}^2$

$\Rightarrow\text{a}^2=\frac{575}{15}-\frac{115}{3}$

Putting $\text{a}^2=\frac{115}{3}$ in equation (ii), we get

$\text{b}^2+16\times\frac{115}{3}-\frac{115}{3}\times\text{b}^2$

$\Rightarrow\text{b}^2-\frac{115}{3}\text{b}^2=-16\times\frac{115}{3}$

$\Rightarrow\frac{3\text{b}^2-115\text{b}^2}{3}=-\frac{16\times115}{3}$

$\Rightarrow-112\text{b}^2=-16\times115$

$\Rightarrow\text{b}^2=\frac{16\times115}{112}$

Substututing the value of a² and b² in (i), we get

$\frac{\frac{\text{x}^2}{115}}{3}+\frac{\frac{\text{y}^2}{115}}{7}=1$

$\Rightarrow\frac{3\text{x}^2+7\text{y}^2}{115}=1$

$\Rightarrow3\text{x}^2+7\text{y}^2=115$

This is the required equation of the ellipse.

 $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where a > b 

$\big[\because\ $axes lie along the coordinates axes $\big]$

Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$

$\Rightarrow\text{b}^2-\text{a}^2\Bigg[1-\bigg(\sqrt{\frac{2}{5}}\bigg)^2\Bigg]$ $\Big[\because\text{e}=\sqrt{\frac{2}{5}}\ \Big]$

$\Rightarrow\text{b}^2=\text{a}^2\Big[1-\frac{2}{5}\Big]$

$\text{b}^2-\text{a}^2\times\frac{3}{5}$

$\Rightarrow\text{b}^2=\frac{3\text{a}^2}{5}\ \dots(\text{ii})$

The required ellipse passes through (-3, 1)

$\therefore\ \frac{(-3)^2}{\text{a}^2}+\frac{1^2}{\text{b}^2}=1$

$\Rightarrow\frac{9}{\text{a}^2}+\frac{1}{\text{b}^2}=1\ \dots(\text{ii})$

Putting $\text{b}^2=\frac{3\text{a}^2}{5}$ in equation (ii), we get

$\frac{9}{\text{a}^2}+\frac{1}{\frac{3\text{a}^2}{5}}=1$

$\Rightarrow\frac{9}{\text{a}^2}+\frac{5}{3\text{a}^2}=1$

$\Rightarrow\frac{1}{\text{a}^2}\Big[\frac{9}{1}+\frac{5}{3}\Big]=1$

$\Rightarrow\frac{27+5}{3}=\text{a}^2$

Putting $\text{a}^2=\frac{32}{3}$ in equation (ii), we get

$\text{b}^2=\frac{3}{5}\times\frac{32}{3}=\frac{32}{5}$

Substituting $\text{a}^2=\frac{32}{3}$ and $\text{b}^2=\frac{32}{5}$ in equation (i), we get

$\frac{\text{x}^2}{\frac{32}{3}}+\frac{\text{y}^2}{\frac{32}{5}}=1$

$\Rightarrow\frac{3\text{x}^2}{32}+\frac{5\text{y}^2}{32}=1$

$\Rightarrow3\text{x}^2+5\text{y}^2=32$

This is the eqution of the required ellipse.

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

$\text{a}=4$

$\Rightarrow\text{a}^2=16$

and, the coordinates of foci are $(\pm3, 0)$ 

$\therefore\ \text{ae}=3$  

$\Rightarrow4\times\text{e}=3$

$\Rightarrow\text{e}=\frac{3}{4}$

Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$

$=4^2\Big[1-\Big(\frac{3}{4}\Big)^2\Big]$

$=16\times\Big(1-\frac{9}{16}\Big)$

$=16\times\frac{7}{16}$

$\frac{\text{x}^2}{16}+\frac{\text{y}^2}{7}=1$ 

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

The coordinates of its vertices and foci are $(\pm\text{a}, 0)$ and $(\pm\text{ae}, 0)$ respectively.

$\therefore\ $ a = 5 and ae = 4 $\Rightarrow\text{e}=\frac{4}{5}$

 $\text{b}^2=\text{a}^2\big(1-\text{e}^2\big)\Rightarrow\text{b}^2=25\Big(1-\frac{16}{25}\Big)=9$

substituting the values of a² and d² in (i), we get

$\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1,$ which is the equation of the required ellipse.

$3\text{x}^2+4\text{y}^2-12\text{x}+8\text{y}+4=0$

$\Rightarrow3\text{x}^2-12\text{x+4}\text{y}^2-8\text{y}+4=0$

$\Rightarrow3\Big(\text{x}^2-4\text{x}\Big)+4\Big(\text{y}^2-2\text{y}\Big)+4=0$

$\Rightarrow3\Big[\text{x}-2\times\text{x}\times2+2^2-2^2\Big]+4\Big[\text{y}^2-2\text{xy}\times1^2-1^2\Big]+4=0$

$\Rightarrow3\Big[\big(\text{x}-2\big)^2-4\Big]+4\Big[\big(\text{y}+1\big)^2-1\Big]+4=0$

$\Rightarrow3\big(\text{x}-2\big)^2-12+4\big(\text{y}-1\big)^2-4+4=0$

$\Rightarrow3\big(\text{x}-2\big)^2+4\big(\text{y}-1\big)^2-12=0$

$\Rightarrow3\big(\text{x}-2\big)^2+4\big(\text{y}-1\big)^2=12$

$\Rightarrow\frac{3\big(\text{x}-2\big)^2}{12}+\frac{4\big(\text{y}-1\big)^2}{{12}}=1$

$\Rightarrow\frac{\big(\text{x}-2\big)^2}{4}+\frac{\big(\text{y}-1\big)^2}{{3}}=1$

$\Rightarrow\frac{\big(\text{x}-2\big)^2}{2^2}+\frac{\big(\text{y}-1\big)^2}{\big(\sqrt{3}\big)^2}=1\ \dots(\text{i})$

$\therefore\ $The coordinates of centre of the ellipse are (2, -1).

x = x + 2 and y = y - 1 $\dots(\text{ii})$

using these relations, equation (i) reduces to 

$\frac{\text{x}^2}{2^2}+\frac{\text{y}^2}{\big(\sqrt{3}\big)^2}=1,$ where $\text{a}=2$ and $\text{b}=\sqrt{3}$

Major-axis = $2\text{a}=2\times2=4$

and, Minor-axis = $2\text{b}=2\times\sqrt{3}=2\sqrt{3}$

Eccentricity: The eccentricity e is given by 

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

$=\sqrt{1-\frac{3}{{4}}}$

$=\sqrt{\frac{1}{4}}$

$=\frac{1}{2}$

foci: The coordinates of the foci with respect to the new axes are given by $\big(\text{x}=\pm\text{ae},\ \text{y}=0\big)$ i.e., $\big(\text{x}=\pm1,\ \text{y}=0\big)$

Putting $\text{x}=\pm\ 1$ and $\text{y}=0$ in equation (iii), we get

$\text{x}=\pm\ 1+2$ and $\text{y}=0+1$

$\Rightarrow\text{x}=2\pm1$ and $\text{y}=1$

So the coordinates of foci with respect to old axes are qiven by $(2\pm1, 1)$ i.e., (3, 1) and (1, 1).

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

$\Rightarrow2\text{a}=26$

$\Rightarrow\text{a}=\frac{26}{2}=13$

$\Rightarrow\text{a}=169$

The coordinates of foci are $(\pm\text{ae, 0}).$ 

$\therefore\ \text{ae}=5$

$\Rightarrow13\times\text{e}=5$

$\Rightarrow\text{e}=\frac{5}{13}$

Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$

$\Rightarrow\text{b}^2=169\Big[1-\Big(\frac{5}{13}\Big)^2\Big]$

$\Rightarrow\text{b}^2=169\Big[1-\frac{25}{169}\Big]$

$\Rightarrow\text{b}^2=169\Big[\frac{144}{169}\Big]$

$\Rightarrow\text{b}^2=144$

$\frac{\text{x}^2}{169}+\frac{\text{y}^2}{144}=1$ 

and  $\text{ae}=4$ $\big[\because\ $foci: $(\pm4, 0)$ $\big]$ 

$=20$

$\text{e}=\frac{2}{3}$ and latus rectum = 5 (Given)

Now, $\frac{2\text{b}^2}{\text{a}}=5$

$\Rightarrow2\text{b}^2=5\text{a}\ \dots(\text{ii})$

$\Rightarrow2\text{a}^2\big(1-\text{e}^2\big)=5\text{a}$ $\Big[\because\ \text{b}^2=\text{a}^2\big(1-\text{e}^2\big)\Big]$

$\Rightarrow2\text{a}^2\Big[1-\frac{4}{9}\Big]=5\text{a}$

$\Rightarrow2\text{a}^2\times\frac{5}{9}=5\text{a}$

$\Rightarrow10\text{a}^2=45\text{a}$

$\Rightarrow\text{a}=\frac{9}{2}$

Substituting the value of a in eq. (ii), we get:

$2\text{b}^2=5\times\frac{9}{2}$

$\Rightarrow\text{b}^2=\frac{45}{4}$

Substituting the value of a² and b² in eq. (i), we get

$\frac{\text{x}^2}{\frac{81}{4}}+\frac{\text{y}^2}{\frac{45}{4}}=1$

$\Rightarrow\frac{4\text{x}^2}{81}+\frac{4\text{y}^2}{45}=1$

This is the required equation of the ellipse.

Using similar triangle principle, we can write

$\frac{\text{Q}}{9}=\frac{\text{y}_1}{3}$

$\text{Q}=3\text{y}_1$

Similarly, $\text{p}=\frac{\text{x}}{3}$

Point $\text{P}(\text{x,y})$

So $\text{OB}=\text{x}+\frac{\text{x}}{3}$

$\text{OA}=\text{y}+3\text{y}=4\text{y}$

Using Pythageorus theorem, we get

$(\text{4y})^2+\Big(\frac{4\text{x}}{3}\Big)^2=12^2$

$\frac{\text{y}^2}{9}+\frac{\text{x}^2}{81}=1$ is the equation of ellipse.

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

The coordinates of the foci are $(\pm2, 0).$ This means that the major and minor axes of the ellipse are along x and y axes respectively and the coordinates of foci are $(\pm\text{ae, 0})$

$\therefore\text{ ae}=2$

$\Rightarrow\text{a}\times\frac{1}{2}=2$ $\big[\because\text{e}=\frac{1}{2}\big]$

$\Rightarrow\text{a}=4$

$\Rightarrow\text{a}^2=16$

Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$

$\Rightarrow\text{b}^2=(4)^2\bigg[1-\big(\frac{1}{2}\big)^2\bigg]$

$\Rightarrow\text{b}^2=16\big[1-\frac{1}{4}\big]$

$\Rightarrow\text{b}^2=16\times\frac{3}{4}=12$

Substituting the value of a² and b² in (i), we get

$\frac{\text{x}^2}{16}+\frac{\text{y}^2}{12}=1$

$\Rightarrow3\text{x}^2+4\text{y}^2=48$

required equation of ellipse.

Here $\text{e}=\frac{1}{2},$ coordinates of S are (-1, 1) and the equation of directrix is x - y + 3 = 0

$\therefore\text{ SP}=\frac{1}{2}\text{PM}$

$\Rightarrow\text{SP}^2=\frac{1}{4}(\text{PM})^2$

$\Rightarrow4\text{SP}^2=\text{PM}^2$

$\Rightarrow4\big[(\text{x}+1)^2+(\text{y}-1)^2\big]=\bigg[\frac{\text{x}-\text{y}+3}{\sqrt{1^2+(-1)^2}}\bigg]^2$

$\Rightarrow4\big[\text{x}^2++1+2\text{x}+\text{y}^2+1-2\text{y}\big]=\frac{(\text{x}-\text{y}+3)^2}{2}$

$\Rightarrow8\big[\text{x}^2+\text{y}^2+2\text{x}-2\text{y}+2\big]=(\text{x}-\text{y}+3)^2$

$\Rightarrow8\text{x}^2+8\text{y}^2+16\text{x}-16\text{y}+16\\=\text{x}^2+(-\text{y})^2+3^2+2\times(-\text{y})\times3+2\times(\text{x})\times(-\text{y})+2\times3\times\text{x}\\$ $\big[\because(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\big]$

$\Rightarrow8\text{x}^2+8\text{y}^2+16\text{x}-16\text{y}+16=\text{x}^2+\text{y}^2+9-6\text{y}-2\text{xy}+6\text{x}$

$\Rightarrow8\text{x}^2-\text{x}^2+8\text{y}^2-\text{y}^2+2\text{xy}+16\text{x}-6\text{x}-16\text{y}+6\text{y}+16-9=0$

$\Rightarrow7\text{x}^2+7\text{y}^2+2\text{xy}+10\text{x}-10\text{y}+7=0$

This is the required equation of the ellipse.

Let S(1, -2) be the focus and ZZ' be the directrix.

Let P(x, y) be any point on the ellipse and let PM be the perpendicular from P on the directix.

Then by the definition of an ellipse, we have:

$\text{SP}=\text{e}.\text{PM},\text{where e}=\frac{1}{2}$

$\Rightarrow\text{SP}^2=\text{e}^2.\text{PM}^2$

$\Rightarrow(\text{x}-1)^2+(\text{y}+2)^2=\Big(\frac{1}{2}\Big)^2\times\Bigg|\frac{\text{3}\text{x}-2\text{y}+5}{\sqrt{(3)^2+(-2)^2}}\Bigg|^2$

$\Rightarrow\text{x}^2+1-2\text{x}+\text{y}^2+4+4\text{y}\\=\Big(\frac{1}{4}\Big)\times\bigg|\frac{9\text{x}^2+4\text{y}^2+25-12\text{xy}-20\text{y}+30\text{x}}{13}\bigg|$ 

$\Rightarrow52\big(\text{x}^2+1-2\text{x}+\text{y}^2+4+4\text{y}\big)\\=9\text{x}^2+4\text{y}^2+25-12\text{xy}-20\text{y}+30\text{x}$ 

$\Rightarrow52\text{x}^2+52-104\text{x}+52\text{y}^2+208+208\text{y}\\=9\text{x}^2+4\text{y}^2+25-12\text{xy}-20\text{y}+30\text{x}$

$\Rightarrow43\text{x}^2+48\text{y}^2-134\text{x}+228\text{y}+235=0$

This is the equation of the required ellipse.

$\text{x}^2+4\text{y}^2-2\text{x}=0$

$\Rightarrow\text{x}^2-2\text{x}+4\text{y}^2=0$

$\Rightarrow\Big(\text{x}^2-2\text{x}+1^2-1^2\Big)+4\text{y}^2=0$

$\Rightarrow\big(\text{x}-1\big)^2-1+4\text{y}^2=0$

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

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

$\Rightarrow\frac{\big(\text{x}-1\big)^2}{1^2}+\frac{\text{y}^2}{\Big(\frac{1}{2}\Big)^2}=1\ \dots(\text{i})$

$\therefore\ $The coordinates of centre of the ellipse are (1, 0).

x = x + 1 and y = y $\dots(\text{ii})$

using these relations, equation (i) reduces to 

$\Rightarrow\frac{\text{x}^2}{1^2}+\frac{\text{y}^2}{\Big(\frac{1}{2}\Big)^2}=1,$ where $\text{a}=1$ and $\text{b}=\frac{1}{2}$

Major-axis = $2\times\text{a}=2\times1=2$

and, Minor-axis = $2\times\text{b}=2\times\frac{1}{2}=1$

Eccentricity: The eccentricity e is given by 

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

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

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

$=\sqrt{\frac{3}{4}}$

$=\frac{\sqrt{3}}{2}$

foci: The coordinates of the foci with respect to the new axes are given by $\big(\text{x}=\pm\text{ae},\ \text{y}=0\big)$ i.e., $\big(\text{x}=\pm\frac{\sqrt{3}}{2},\ \text{y}=0\big)$

Putting $\text{x}=\pm\frac{\sqrt{3}}{2}$ and $\text{y}=0$ in equation (iii), we get

$\text{x}=\pm\frac{\sqrt{3}}{2}+1$ and $\text{y}=0$

$\text{x}=1\pm\frac{\sqrt{3}}{2}$ and $\text{y}=0$

So, the coordinates of foci with respect to the old axes given by $\Big(1\pm\frac{\sqrt{3}}{2},\ 0\Big).$

$4\text{x}^2+\text{y}^2-8\text{x}+2\text{y}+1=0$

$\Rightarrow4\big(\text{x}^2-2\text{x}\big)+\big(\text{y}^2+2\text{y}\big)+1=0$

$\Rightarrow4\Big[\big(\text{x}^2-2\text{x}+1\big)-1\Big]+\Big[\big(\text{y}^2+2\text{y}-1\big)\Big]+1=0$

$\Rightarrow4\Big[\big(\text{x}-1\big)^2-1\Big]+\Big[\big(\text{y}+1\big)^2-1\Big]+1=0$

$\Rightarrow4\big(\text{x}-1\big)^2-4+\big(\text{y}+1\big)^2-1+1=0$

$\Rightarrow4\big(\text{x}-1\big)^2+\big(\text{y}+1\big)^2-4=0$

$\Rightarrow4\big(\text{x}-1\big)^2+\big(\text{y}+1\big)^2=4$

$\Rightarrow\frac{\big(\text{x}-2\big)^2}{1}+\frac{\big(\text{y}+3\big)^2}{4}=1$

$\Rightarrow\frac{\big(\text{x}-1\big)^2}{1^2}+\frac{\big(\text{y}+1\big)^2}{2^2}=1\ \dots(\text{i})$

$\therefore\ $The coordinates of centre of the ellipse are (1, -1).

Shifting the origin at (1, -1) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by x and y, we have

x = x + 1 and y = y - 1  $\dots(\text{ii})$

using these relations, equation (i) reduces to 

$\frac{\text{x}^2}{\text{1}^2}+\frac{\text{y}^2}{\text{2}^2}=1$

This is of the form

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where a = 1 and b = 2

clearly, a > b. so, the given equation represents an ellipse whose and minor axes are along x and y axes respectively.

Length of the axes:

Major-axis = $2\text{a}=2\times2=4$

Minor-axis = $2\text{a}=2\times1=2$

Eccentricity: The eccentricity e is given by 

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

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

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

$=\sqrt{\frac{3}{4}}$

$={\frac{\sqrt3}{2}}.$

foci: The coordinates of the foci with respect to the new axes are given by $\big(\text{x}=0,\ \text{y}=\pm\text{be}\big)$ i.e., $\Big(\text{x}=0,\ \text{y}=\pm\sqrt{3}\Big)$

Putting $\text{x}=0$ and $\text{y}=\pm\sqrt{3}$ in equation (iii), we get

$\text{x}=0+1$ and $\text{y}=\pm\sqrt{3}-1$

$\Rightarrow\text{x}=1$ and $\text{y}=-1\pm\sqrt{3}$

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

$\Rightarrow2\text{a}=16$

$\Rightarrow\text{a}=\frac{16}{2}=8$

$\Rightarrow\text{a}=64$

The coordinates of foci are $(0, \pm\text{be}).$ 

$\therefore\ \text{ae}=6$ $\big[\therefore\ $foci: $(0, \pm6)\ \big]$ 

$\Rightarrow(\text{be})^2=36$

Now, $\text{a}^2=\text{b}^2(1-\text{e}^2)$

$\Rightarrow\text{a}^2=\text{b}^2-\text{b}^2\text{e}^2$

$\Rightarrow64=\text{b}^2-36$ 

$\big[\because(\text{be}^2)=36$ and $\text{a}^2=64\big]$

$\Rightarrow64+36=\text{b}^2$

$\Rightarrow\text{b}^2=100$

$\frac{\text{x}^2}{64}+\frac{\text{y}^2}{100}=1$ 

$\text{x}^2+4\text{y}^2-4\text{x}+24\text{y}+31=0$

$\Rightarrow\text{x}^2+4\text{x}+4+\big(\text{y}^2+6\text{y}\big)+31=0$

$\Rightarrow\big[\text{x}^2-2\times\text{x}\times2^2-2^2\big]+4\big[\text{y}^2+2\times3\times\text{xy}+3^2-3^2\big]+31=0$

$\Rightarrow\Big[\big(\text{x}-2\big)^2-2^2\Big]+4\Big[\big(\text{y}+3\big)^2-9\Big]+31=0$

$\Rightarrow\big(\text{x}-2\big)^2-4+4\big(\text{y}+3\big)^2-36+31=0$

$\Rightarrow\big(\text{x}-2\big)^2+4\big(\text{y}+3\big)^2-5-4=0$

$\Rightarrow\big(\text{x}-2\big)^2+4\big(\text{y}+3\big)^2=9$

$\Rightarrow\frac{\big(\text{x}-1\big)^2}{9}\frac{4\big(\text{y}+3\big)^2}{9}=1$

$\Rightarrow\frac{\big(\text{x}-2\big)^2}{9}+\frac{\big(\text{y}+3\big)^2}{\frac{9}{4}}=1$

$\Rightarrow\frac{\big(\text{x}-2\big)^2}{(3)^2}+\frac{\big(\text{y}+3\big)^2}{\Big(\frac{3}{{2}}\Big)^2}=1\ \dots(\text{i})$

$\therefore\ $The coordinates of centre of the ellipse are (2, -3).

x = x + 1 and y = y - 3  $\dots(\text{ii})$

using these relations, equation (i) reduces to 

$\Rightarrow\frac{\text{x}^2}{3^2}+\frac{\text{y}^2}{\Big(\frac{3}{{2}}\Big)^2}=1\ \dots(\text{iii})$

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where a = 3

and $\text{b}=\frac{3}{{2}}.$

major-axis = $2\text{a}=2\times3=6$

and, minor-axis = $2\text{b}=2\times\frac{3}{{2}}=3$

eccentricity: The eccentricity e is given by 

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

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

$=\sqrt{1-\frac{9}{4\times9}}$

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

$=\sqrt{1-\frac{3}{4}}$

$=\frac{\sqrt{3}}{2}.$

foci: The coordinates of the foci with respect to the new axes are given by $\big(\text{x}=\pm\text{ae},\text{y}=0\big)$ i.e., $\Big(\text{x}=\pm\frac{3\sqrt{3}}{2},\ \text{y}=0\Big)$

Putting $\text{x}=\pm\frac{3\sqrt{3}}{2}$ and $\text{y}=0$ in equation (ii), we get

$\text{x}=\pm2$ and $\text{y}=0-3$

$\Rightarrow\text{x}=2\pm\frac{3\sqrt{3}}{\sqrt{2}}$ and $\text{y}=-3.$

So, the  coordinates of foci with respect to old axes are given by $\bigg(2\pm\frac{3\sqrt{3}}{\sqrt{2}},-3\bigg)$

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

The coordinates of foci are $(+\text{ae}, 0)$ and $(-\text{ae, 0}).$

$\therefore\ $a = 4 $\big[\because\ $focai: $(\pm4, 0)$ $\big]$

$\Rightarrow\text{a}\times\frac{1}{3}=4$ $\Big[\because\text{e}=\frac{1}{3}\Big]$

$\Rightarrow\text{a}=12$

$\Rightarrow\text{a}^2=144$  

Now, $\text{b}^2=\text{a}^2\big(1-\text{e}^2\big)$

$\Rightarrow\text{b}^2=144\Big[1-\Big(\frac{1}{3}\Big)^2\Big]$

$\Rightarrow\text{b}^2=144\Big[1-\frac{1}{9}\Big]$

$\Rightarrow\text{b}^2=144\times\frac{8}{9}$

$\Rightarrow\text{b}^2=16\times8=128$

$=\frac{\text{x}^2}{144}+\frac{\text{y}^2}{128}=1$ 

$\Rightarrow\frac{1}{16}\Big[\frac{\text{x}^2}{9}+\frac{\text{y}^2}{8}\Big]=1$

$\Rightarrow\frac{\text{x}^2}{9}+\frac{\text{y}^2}{8}=16$

From above figure,

Assum e length AB = l 

AP = a, PB = b

Assume $\overbrace{\text{ABQ}}=\theta$

So $\text{x}_1=\text{a}\cos\theta,\ \text{y}_1=\text{b}\sin\theta$

$\Rightarrow\Big(\frac{\text{x}_1}{\text{a}}\Big)^2+\Big(\frac{\text{y}_1}{\text{b}}\Big)^2=1$

$4\text{x}^2+16\text{y}^2-24\text{x}-32\text{y}-12=0$

$\Rightarrow4\text{x}^2-24\text{x}++16\text{y}^2-32\text{y}-12=0$

$\Rightarrow4\Big(\text{x}^2-6\text{x}\Big)+16\Big(\text{y}^2-2\text{y}\Big)-12=0$

$\Rightarrow4\Big[\text{x}^2-2\times\text{x}\times3+3^2-3^2\Big]+16\Big[\text{y}^2-2\text{y}\times1^2-1^2\Big]-12=0$

$\Rightarrow4\Big[\big(\text{x}-3\big)^2-9\Big]+16\Big[\big(\text{y}-1\big)^2-1\Big]-12=0$

$\Rightarrow4\big(\text{x}-3\big)^2-36+16\big(\text{y}-1\big)^2-16-12=0$

$\Rightarrow4\big(\text{x}-3\big)^2+16\big(\text{y}-1\big)^2-36-28=0$

$\Rightarrow4\big(\text{x}-3\big)^2+16\big(\text{y}-1\big)^2-64=0$

$\Rightarrow4\big(\text{x}-3\big)^2+16\big(\text{y}-1\big)^2=64$

$\Rightarrow\frac{4\big(\text{x}-3\big)^2}{64}+\frac{\big(\text{y}-1\big)^2}{{64}}=1$

$\Rightarrow\frac{\big(\text{x}-3\big)^2}{16}+\frac{\big(\text{y}-1\big)^2}{{4}}=1$

$\Rightarrow\frac{\big(\text{x}-3\big)^2}{(4)^2}+\frac{\big(\text{y}-1\big)^2}{(2)^2}=1\ \dots(\text{i})$

$\therefore\ $The coordinates of centre of the ellipse are (3, 1).

x = x + 3 and y = y + 1 $\dots(\text{ii})$

using these relations, equation (i) reduces to 

$\frac{\text{x}^2}{4^2}+\frac{\text{y}^2}{2^2}=1,$ where $\text{a}=4$ and $\text{b}=2$

Major-axis = $2\text{a}=2\times4=8$

and, Minor-axis = $2\text{b}=2\times2=4$

Eccentricity: The eccentricity e is given by 

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

$=\sqrt{1-\frac{4}{{16}}}$

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

$=\sqrt{\frac{3}{4}}$

$=\frac{\sqrt{3}}{2}$

foci: The coordinates of the foci with respect to the new axes are given by $\big(\text{x}=\pm\text{ae},\ \text{y}=0\big)$ i.e., $\big(\text{x}=\pm2\sqrt{3},\ \text{y}=0\big)$

Putting $\text{x}=\pm2\sqrt{3}$ and $\text{y}=0$ in equation (iii), we get

$\text{x}=\pm2\sqrt{3}+3$ and $\text{y}=0+1$

$\Rightarrow\text{x}=3\pm2\sqrt{3}$ and $\text{y}=1$

Then ae = 3

Also x = 4 and y = 1 [Ellipse passing through (4, 1)]

Substititing the values of x and y in eq. (i), we get:

$\frac{4^2}{\text{a}^2}+\frac{1^2}{\text{b}^2}=1$

$\Rightarrow\frac{16}{\text{a}^2}+\frac{1}{\text{b}^2}=1$

Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$

$\Rightarrow\text{b}^2=\text{a}^2-\text{a}^2\text{e}^2$

$\Rightarrow\text{b}^2=\text{a}^2-9$ or $\text{a}^2=\text{b}^2+9\ \dots(\text{ii})$

$\Rightarrow\frac{16}{\text{a}^2}+\frac{1}{\text{b}^2}=1$

$\Rightarrow16\text{b}^2+\text{a}^2+\text{a}^2\text{b}^2$

$\Rightarrow16\text{b}^2+\text{b}^2+9=\text{b}^2\big(\text{b}^2+9\big)$

$\Rightarrow17\text{b}^2+9=\text{b}^4+9\text{b}^2$

$\Rightarrow\text{b}^4-8\text{b}^2-9=0$

$\Rightarrow\big(\text{b}^2-9\big)\big(\text{b}^2+1\big)$

$\Rightarrow\text{b}=\pm3$

Substituting the value of b in eq. (2), we get:

$\text{a}=3\sqrt{2}$

$\therefore\frac{\text{x}^2}{18}+\frac{\text{y}^2}{9}=1$

This is the required equation of the ellipse.

$\frac{(\text{x}+2)^2}{\text{a}^2}+\frac{(\text{y}-3)^2}{\text{b}^2}=1$ $\big[\because\ $centre: (-2, 3) $\dots(\text{i})\ \big]$

$\Rightarrow\text{a}^2=9$

$\Rightarrow\text{b}^2=4$

$\frac{(\text{x}+2)^2}{9}+\frac{(\text{y}-3)^2}{4}=1$

$\Rightarrow\frac{4(\text{x}+2)^2+9(\text{y}-3)^2}{36}=1$

$\Rightarrow4(\text{x}+2)^2+9(\text{y}-3)^2=36$

$\Rightarrow\big[\text{x}^2+4+4\text{x}\big]+9\big[\text{y}^2+9-6\text{y}\big]=36$

$\Rightarrow4\text{x}^2+16+16\text{x}^2+9\text{y}^2+81-54\text{y}=36$

$\Rightarrow4\text{x}^2+9\text{y}^2+16\text{x}-54\text{y}+16+81-36=0$

$\Rightarrow4\text{x}^2+9\text{y}^2+16\text{x}-54\text{y}+61=0$

Here $\text{e}-\frac{1}{2}.$ coordinates of S are (0, 1) and the euation of the directrix is x + y - 0.

$\therefore\text{SP}=\frac{1}{2}\text{PM}$

$\Rightarrow\text{SP}^2=\frac{1}{2}(\text{PM})^2$

$\Rightarrow4\text{SP}^2-(\text{PM})^2$

$\Rightarrow4\big[(\text{x-0})^2+(\text{y-1})^2\big]-\bigg[\frac{\text{x+y}}{\sqrt{1^2+1^2}}\bigg]$

$\Rightarrow4\big[\text{x}^2+\text{y}^2+1-2\text{y}\big]=\frac{(\text{x+h})^2}{2}$

$\Rightarrow4\times2\big[\text{x}^2+\text{y}^2-2\text{y}+1\big]=\text{x}^2+\text{y}^2+2\text{xy}$

$\Rightarrow8\text{x}^2+8\text{y}^2-16\text{y}+8=\text{x}^2+ \text{y}^2+2\text{xy}$

$\Rightarrow8\text{x}^2-\text{x}^2+8\text{y}^2-2\text{xy}-16\text{y}+8=0$

$\Rightarrow7\text{x}^2+7\text{y}^2-2\text{xy}-16\text{y}+8-0$

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

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

$\Rightarrow\sqrt\frac{2}{5}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$ $\bigg[\because\text{eccentricity}=\sqrt{\frac{2}{5}}\bigg]$

$\Rightarrow\sqrt\frac{2}{5}=1-\frac{\text{b}^2}{\text{a}^2}$

$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=1-\frac{2}{5}$

$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\frac{3}{5}$

$\Rightarrow5\text{b}^2=3\text{a}^2$

$\Rightarrow\text{b}^2=\frac{3\text{a}^2}{5}\ \dots(\text{ii})$

Putting the value of $\text{b}^2=\frac{3\text{a}^2}{5}$in equation (ii), we get

$\frac{9}{\text{a}^2}+\frac{1}{\frac{3\text{a}^2}{5}}=1$

$\Rightarrow\frac{9}{\text{a}^2}+\frac{5}{3\text{a}^2}=1$

$\Rightarrow\frac{1}{\text{a}^2}\big[9+\frac{5}{3}\big]=1$

$\Rightarrow9+\frac{5}{3}=\text{a}^2$

$\Rightarrow\text{a}^2=\frac{32}{3}\ \dots(\text{iii})$

Putting $\text{a}^2=\frac{32}{3}$ in equation (ii), we get

$\text{b}^2=\frac{3}{5}\times\frac{32}{3}=\frac{32}{5}\ \dots(\text{iv})$

$\therefore$ The required equation of ellipse is

$\frac{\text{x}^2}{\frac{32}{3}}+\frac{\text{y}^2}{\frac{32}{5}}=1$

$\Rightarrow\frac{3\text{x}^2}{32}+\frac{5\text{y}^2}{32}=1$

$\Rightarrow3\text{x}^2+5\text{y}^2=32.$

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ \dots(\text{i})$

$\therefore\ \frac{2\text{b}^2}{\text{a}}=5$

$\Rightarrow\text{b}^2=\frac{5\text{a}}{2}\ \dots(\text{ii})$

Now, $\text{b}^2=\text{a}^2\big(1-\text{e}^2\big)$

$\Rightarrow\frac{5\text{a}}{2}=\text{a}^2\Big[1-\big(\frac{2}{3}\big)^2\Big]$ $\big[\because\text{e}=\frac{2}{3}\big]$

$\Rightarrow\frac{5\text{a}}{2}=\text{a}^2\Big[1-\frac{4}{9}\Big]$

$\Rightarrow\frac{5\text{a}}{2}=\text{a}\Big(\frac{5}{9}\Big)$

$\Rightarrow\frac{5}{2}\times\frac{9}{5}=\text{a}$

$\Rightarrow\text{a}=\frac{9}{2}$

$\Rightarrow\text{a}^2=\frac{81}{4}$

Putting $\text{a}=\frac{9}{2}$ in $\text{b}^2\frac{5\text{a}}{2},$ we get

$\text{b}^2=\frac{5}{2}\times\frac{9}{2}$

$\Rightarrow\text{b}^2=\frac{45}{4}$

Substituting $\text{a}^2=\frac{81}{4}$ and $\text{b}^2=\frac{45}{4}$ in equation (i), we get

$\frac{\text{x}^2}{\frac{81}{4}}+\frac{\text{y}^2}{\frac{45}{4}}=1$

$\Rightarrow\frac{4\text{x}^2}{81}=\frac{4\text{y}^2}{45}=1$

$\frac{4\text{x}^2\times5+4\text{y}^2\times9}{405}=1$

$\Rightarrow20\text{x}^2+36\text{y}^2=405$

 $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where a > b $\dots(\text{i})$

$\therefore\ \frac{(4)^2}{\text{a}^2}+\frac{(3)^2}{\text{b}^2}=1$

$\Rightarrow\frac{16}{\text{a}^2}+\frac{9}{\text{b}^2}=1$

$\Rightarrow16\text{b}^2+9\text{a}^2=\text{a}^2\text{b}^2\ \dots(\text{ii})$

and $\frac{(-1)^2}{\text{a}^2}+\frac{(4)^2}{\text{b}^2}=1$

$\Rightarrow\frac{1}{\text{a}^2}\times\frac{16}{\text{b}^2}=1$

$\Rightarrow\text{b}^2+16\text{a}^2=\text{a}^2\text{b}^2\ \dots(\text{iii})$

$16\text{b}^2+256\text{a}^2=16\text{a}^2\text{b}^2\ \dots(\text{iv})$

$256\text{a}^2-9\text{a}^2-16\text{a}^2\text{b}^2-\text{a}^2\text{b}^2$

$\Rightarrow247\text{a}^2=15\text{a}^2\text{b}^2$

$\Rightarrow\frac{247}{15}-\text{b}^2$

$\Rightarrow\text{b}^2-\frac{247}{15}$

Putting $\text{b}^2-\frac{247}{15}$ in equation (iii) we get

$\frac{247}{15}+16\text{a}^2=\text{a}^2\times\frac{247}{15}$

$\Rightarrow16\text{a}^2-\frac{247\text{a}^2}{15}=\frac{-247}{15}$

$\Rightarrow\frac{240\text{a}^2-247\text{a}^2}{15}=\frac{-247}{15}$

$\Rightarrow-7\text{a}^2=-247$

$\Rightarrow\text{a}^2=\frac{247}{7}$

Putting $\text{a}^2=\frac{247}{7}$ and $\text{b}^2=\frac{247}{15}$ in equation equation (i), we get

$\frac{\text{x}^2}{\frac{247}{7}}+\frac{\text{y}^2}{\frac{247}{15}}=1$

$\Rightarrow\frac{7\text{x}^2}{247}+\frac{15\text{y}^2}{247}=1$

Now, $2\text{a}=12$ $\big[\because$ major axis = 12$\big]$

$\text{x}^2+2\text{y}^2-2\text{x}+12\text{y}+10=0$

$\Rightarrow\text{x}^2-2\text{x}+2\text{y}^2+12\text{y}+10=0$

$\Rightarrow\big(\text{x}^2-2\text{x}+1-1\big)+2\big(\text{y}^2+6\text{y}\big)+10=0$

$\Rightarrow\Big[\big(\text{x}-1\big)^2-1\Big]+2\Big[\big(\text{y}^2+2\times\text{y}\times3+9\big)\Big]+10=0$

$\Rightarrow\big(\text{x}-1\big)^2-1+2\Big[\big(\text{y}+3\big)^2-9\Big]+10=0$

$\Rightarrow\big(\text{x}-1\big)^2+2\big(\text{y}+3\big)^2-18-1+10=0$

$\Rightarrow\big(\text{x}-1\big)^2+2\big(\text{y}+3\big)^2-19+10=0$

$\Rightarrow\big(\text{x}-1\big)^2+2\big(\text{y}+3\big)^2-9=0$

$\Rightarrow\big(\text{x}-1\big)^2+2\big(\text{y}+3\big)^2=9$

$\Rightarrow\frac{\big(\text{x}-1\big)^2}{9}+2\frac{\big(\text{y}+3\big)^2}{9}=1$

$\Rightarrow\frac{\big(\text{x}-1\big)^2}{9}+\frac{\big(\text{y}+3\big)^2}{\frac{9}{2}}=1$

$\Rightarrow\frac{\big(\text{x}-1\big)^2}{(3)^2}+\frac{\big(\text{y}+3\big)^2}{\Big(\frac{3}{\sqrt{2}}\Big)^2}=1\ \dots(\text{i})$

$\therefore\ $The coordinates of centre of the ellipse are (1, -3).

x = x + 1 and y = y - 3  $\dots(\text{i})$

using these relations, equation (i) reduces to 

$\Rightarrow\frac{\text{x}^2}{3^2}+\frac{\text{y}^2}{\Big(\frac{3}{\sqrt{2}}\Big)^2}=1\ \dots(\text{iii})$

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where a = 3

and $\text{b}=\frac{3}{\sqrt{2}}.$

major-axis = $2\text{a}=2\times3=6$

and, minor-axis = $2\text{b}=\frac{2\times3}{\sqrt{2}}=3\sqrt{2}$

eccentricity: The eccentricity e is given by 

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

$=\sqrt{1-\frac{\Big(\frac{3}{\sqrt{2}}\Big)^2}{3^2}}$

$\sqrt{1-\frac{9}{2\times9}}$

$=\sqrt{1-\frac{1}{2}}$

$=\sqrt{\frac{1}{2}}$

$={\frac{1}{2}}$

foci: The coordinates of the foci with respect to the new axes given by $\big(\text{x}=\pm\text{ae},\text{y}=0\big)$ i.e., $\Big(\text{x}=\pm\frac{3}{\sqrt{2}},\ \text{y}=0\Big)$

Putting $\text{x}=\pm\frac{3}{\sqrt{2}}$ and $\text{y}=0$ in equation (ii), we get

$\text{x}=\pm\frac{3}{\sqrt{2}}+1$ and $\text{y}=0-3$

$\Rightarrow\text{x}=1\pm\frac{3}{\sqrt{2}}$ and $\text{y}=-3$

$\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ ,\dots(\text{i})$

The coordinates of vertices are $(0,\pm\text{b})$ i.e., $(0,\pm10).$

$\therefore\ \text{b}=10$

$\Rightarrow\text{b}^2=100$

Now, $\text{a}^2=\text{b}^2\big(1-\text{e}^2\big)$

$\Rightarrow\text{a}^2=100\Big[1-\Big(\frac{4}{5}\Big)^2\Big]$

$\Rightarrow\text{a}^2=100\Big[1-\frac{16}{25}\Big]$

$\Rightarrow\text{a}^2=100\Big[\frac{9}{25}\Big]$

$\Rightarrow\text{a}^2=4\times9=36$

$\frac{\text{x}^2}{36}+\frac{\text{y}^2}{100}=1$

$\Rightarrow\frac{100\text{x}^2+36\text{y}^2}{3600}=1$

$\Rightarrow100\text{x}^2+36\text{y}^2=3600$

i. e. 2ae = 8 $\dots(1)$

i. e. $\frac{2\text{a}}{\text{e}}=\ 18\dots(2)$

$\text{e}=\frac{8}{2\text{a}}$

$\frac{2\text{a}}{\frac{8}{2\text{a}}}=18$

$\Rightarrow4\text{a}^2=18\times8$

$\Rightarrow\text{a}^2=36$ or $\text{a}=6$

Now, $2\text{ae}=8$

$\Rightarrow12\text{e}=8$ or $\text{a}^2\big(1-\text{e}^2\big)$

$\Rightarrow\text{b}^2=36\big(1-\frac{4}{9}\big)$

$\Rightarrow\text{b}^2=36\times\frac{5}{9}$

$\Rightarrow\text{b}^2=20$

$\therefore\ \frac{\text{x}^2}{36}+\frac{\text{y}^2}{20}=1$

 $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ where a < b $\dots(\text{i})$

Now, $\text{a}^2=\text{b}^2\big(1-\text{e}^2\big)$

$\Rightarrow\text{a}^2=\text{b}^2\Big[1-\Big(\frac{3}{4}\Big)^2\Big]$

 $\Rightarrow\text{a}^2=\text{b}^2\Big[1-\frac{9}{16}\Big]$

$\Rightarrow\text{a}^2=\text{b}^2\times\frac{7}{16}$

$\Rightarrow\text{a}^2=\frac{7}{16}\text{b}^2\ \dots(\text{ii})$

$\therefore\ \frac{(6)^2}{\text{a}^2}+\frac{(4)^2}{\text{b}^2}=1$

$\Rightarrow\frac{36}{\text{a}^2}+\frac{16}{\text{b}^2}=1$

$\Rightarrow\frac{36}{\frac{7}{16}\text{b}^2}+\frac{16}{\text{b}^2}=1$ $\Big[\because\ \text{a}^2=\frac{7}{16}\text{b}^2\Big]$

$\Rightarrow\frac{36\times16}{7\text{b}^2}+\frac{16}{\text{b}^2}=1$

$\Rightarrow\frac{576}{7\text{b}^2}+\frac{16}{\text{b}^2}=1$

$\Rightarrow\frac{36}{\frac{7}{16}\text{b}^2}+\frac{16}{\text{b}^2}=1$ $\Big[\because\text{a}^2=\frac{7}{16}\text{b}^2\Big]$

$\Rightarrow\frac{36\times16}{7\text{b}^2}+\frac{16}{\text{b}^2}=1$

$\Rightarrow\frac{576}{7\text{b}^2}+\frac{16}{\text{b}^2}=1$

$\Rightarrow\frac{1}{\text{b}^2}\Big[\frac{576}{7}+\frac{16}{1}\Big]=1$

$\Rightarrow\frac{576}{7}+\frac{16}{1}=\text{b}^2$

$\Rightarrow\frac{576+112}{7}=\text{b}^2$

$\Rightarrow\text{b}^2=\frac{688}{7}.$

Putting $\text{b}^2=\frac{688}{7}$ in equation (i), we get

$\text{a}^2=\frac{7}{16}\times\frac{688}{7}$

$\Rightarrow\text{a}^2=\frac{688}{16}=43$

Putting $\text{a}^2=43$ and $\text{b}^2=\frac{688}{7}$ in equation (i), we get

$\frac{\text{x}^2}{43}+\frac{\text{y}^2}{\frac{688}{7}}=1$

$\Rightarrow\frac{\text{x}^2}{43}+\frac{7\text{y}^2}{688}=1$

This is the equation of the required ellipse.