Solution : Here equation of ellipse is

$\frac{\mathrm{x}^{2}}{10-\mathrm{a}}+\frac{\mathrm{y}^{2}}{4-\mathrm{a}}=1$

$\mathrm{a}^{2}=10-\mathrm{a}$ and $\mathrm{b}^{2}=4-\mathrm{a}$

$\therefore 10-\mathrm{a}>0$ and $4-\mathrm{a}>0$

$10>\mathrm{a}$ and $4>\mathrm{a}$

$\therefore \mathrm{a}<4$.

Solution :

$ \begin{aligned} & \frac{x^{2}}{16}+\frac{y^{2}}{9}=1 \\ & a^{2}=16 b^{2}=9 \\ & a=4 \quad b=3 \quad a>b \\ & \therefore e=\sqrt{1-\frac{b^{2}}{a^{2}}} \\ & =\sqrt{1-\frac{9}{16}} \\ & =\sqrt{\frac{7}{16}} \\ & =\frac{\sqrt{7}}{4} \end{aligned} $

Foci $=( \pm ae, 0) \therefore$ foci is $( \pm \sqrt{7}, 0)$

Radius of the circle through foci & centre $(0,3)$ is $\sqrt{(\sqrt{7})^{2}+3^{2}}=\sqrt{7+9}=\sqrt{16}=4$.

Solution : Let $\mathrm{P}(\mathrm{x} y)$ be any point on the ellipse

Then by definition $\mathrm{SP}=\mathrm{ePM}$.

$\sqrt{(\mathrm{x}-1)^{2}+(\mathrm{y}+1)^{2}}=\frac{1}{2}\left|\frac{\mathrm{x}-\mathrm{y}-3}{\sqrt{2}}\right|$

$(x-1)^{2}+(y+1)^{2}=\frac{1}{8}\left(x^{2}+y^{2}+9-2 x y+6 y-6 x\right)$

$8 x^{2}-16 x+8+8 y^{2}+16 y+8=x^{2}+y^{2}-2 x y+6 y-6 x+9$

$7 x^{2}+7 y^{2}+2 x y-10 x+10 y+7=0$

Solution :

$(5 x-1)^{2}+(5 y-2)^{2}=\left(\lambda^{2}-\lambda+1\right)(3 x+4 y-1)^{2}$

$25\left(\mathrm{x}-\frac{1}{5}\right)^{2}+25\left(\mathrm{y}-\frac{2}{5}\right)^{2}=(\lambda-1)^{2}(3 \mathrm{x}+4 \mathrm{y}-1)^{2}$

$\left(\mathrm{x}-\frac{1}{5}\right)^{2}+\left(\mathrm{y}-\frac{2}{5}\right)^{2}=(\lambda-1)^{2} \frac{(3 \mathrm{x}+4 \mathrm{y}-1)^{2}}{25}$

$\mathrm{Sp}^{2}=\mathrm{e}^{2}(\mathrm{PM})^{2}$

$\therefore \mathrm{e}=|\lambda-1|$

In ellipse $0<\mathrm{e}<1$

$ \begin{aligned} & \therefore 0<|\lambda-1|<1 \\ & 0<\lambda<2-\{1\} \\ & \therefore \lambda \in(0,2)-\{1\} \end{aligned} $

Solution:

$ \begin{array}{ll} \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1 \quad \mathrm{a}>\mathrm{b} \\ \text { major axis } & =2 \mathrm{a} \\ \text { latus rectum is } & =\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\mathrm{a} \end{array} $

$ \begin{aligned} & \text { According to question } ; \frac{2 b^{2}}{a}=a \\ & 2 b^{2}=a^{2} \\ & \text { eccentricity e }=\sqrt{1-\frac{b^{2}}{a^{2}}} \\ & =\sqrt{1-\frac{1}{2}} \\ & =\frac{1}{\sqrt{2}} \end{aligned} $

Solution :

$\mathrm{S}(5,12), \mathrm{S}^{\prime}(24,7)$

$ \begin{aligned} & \mathrm{SP}=\sqrt{5^{2}+12^{2}}=13 \\ & \begin{aligned} & \mathrm{S}^{\prime} \mathrm{P}=\sqrt{24^{2}+7^{2}}=25 \\ & \mathrm{~S}^{\prime}=2 \mathrm{ae}=\sqrt{19^{2}+5^{2}} \\ &=\sqrt{361+25} \\ &=\sqrt{386} \\ &=\frac{\sqrt{386}}{2} \\ & \mathrm{ae} \end{aligned} \\ & \begin{aligned} \mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P} & =2 \mathrm{a}=13+25 \\ 2 \mathrm{a} & =38 \\ \mathrm{a} & =19 \\ \mathrm{e} & =\frac{\mathrm{SS}^{\prime}}{\mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P}}=\frac{2 \mathrm{ae}}{2 \mathrm{a}} \\ & =\frac{\sqrt{386}}{38} \end{aligned} \end{aligned} $

Solution :

Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be a point divides double ordinate in the ratio $1: 2$ internally

Let coordinates of ends of double ordinate $\left(h, y _{1}\right)$ and $\left(h,-y _{1}\right)$.

$\therefore$ By section formula $\mathrm{k}=\frac{-\mathrm{y} _{1}+2 \mathrm{y} _{1}}{3}=\frac{\mathrm{y} _{1}}{3}$

$ \mathrm{y} _{1}=3 \mathrm{k} $

Now the point $\left(\mathrm{h}, \mathrm{y} _{1}\right)=(\mathrm{h}, 3 \mathrm{k})$ lies on the ellipse

$\therefore \frac{\mathrm{h}^{2}}{\mathrm{a}^{2}}+\frac{9 \mathrm{k}^{2}}{\mathrm{~b}^{2}}=1$ or $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{9 \mathrm{y}^{2}}{\mathrm{~b}^{2}}=1 \quad(\because(\mathrm{h}, \mathrm{k})$ are arbitrary $)$

Solution : Here equation of ellipse is

$9 x^{2}+16 y^{2}=144$

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$

$16>9 \therefore a>b$

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

$=\sqrt{1-\frac{9}{16}}$

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

$\mathrm{e}=\frac{\sqrt{7}}{4}$

Foci is $( \pm \mathrm{ae}, 0)=( \pm \sqrt{7}, 0)$

Semi major axis is $=\frac{2 \mathrm{a}}{2}=\mathrm{a}=4$

$\mathrm{CS}=\sqrt{7}$ and semi major axis is 4.

$\therefore$ Required ratio is $\sqrt{7}: 4$.