(4)

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be the given ellipse with length of minor axis as 2b and distance between foci is 2ae.

$\therefore 2b=\frac{1}{4}\times \left(2ae\right) \Rightarrow 4b=ae \Rightarrow \frac{b}{a}=\frac{e}{4}$          ... (i)

We know, $e=\sqrt{1–\frac{b^2}{a^2}} \Rightarrow e=\sqrt{1–\frac{e^2}{16}}$          [From (i)]

$\Rightarrow e^2=\frac{16–e^2}{16} \Rightarrow 16e^2=16–e^2 \Rightarrow 17e^2=16$

$e=\frac{4}{\sqrt{17}}$.

(1)

$\frac{x–\sqrt{5}}{\cos \theta }=\frac{y–\sqrt{5}}{\sin \theta }=r$

$x=r \cos \theta +\sqrt{5}$

$y=r \sin \theta +\sqrt{5}$

$25x^2+36y^2=900, (x,y)$ lie on E

$r^2[25 {\cos }^2\theta +36 {\sin }^2\theta ]+r[50\sqrt{5}\cos \theta +72\sqrt{5}\sin \theta ]+25\left[5\right]+36\left[5\right]=900$

$r_1{r}_2=\frac{900–305}{25+11 {\sin }^2\theta }, {\left(r_1{r}_2\right)}_{\max }=\frac{595}{25}=\frac{119}{5}$

${|r_1·r_2|}_{\max }=\left(\frac{595}{25}\right)=\left(\frac{119}{5}\right) \mathrm{at} \theta =0$

$\frac{x^2}{36}+\frac{y^2}{25}=1$

At $y=\sqrt{5}$

$\frac{x^2}{36}+\frac{1}{5}=1 \Rightarrow \frac{x^2}{36}=\frac{4}{5} \Rightarrow x=\pm \frac{12}{\sqrt{5}}$

$P{A}^2+P{B}^2={(PA+PB)}^2–2PA·PB={\left(\frac{24}{\sqrt{5}}\right)}^2–2·\frac{119}{5}$

$5(P{A}^2+P{B}^2)=5(\frac{{24}^2}{5}–\frac{2.119}{5})={24}^2–238=338$.

(4)

Given, foci $=(\pm 2,0)=(\pm ae,0)$ and eccentricity $\left(e\right)=\frac{1}{2}$

$\therefore ae=2$

$\Rightarrow a\times \frac{1}{2}=2 \Rightarrow a=4$                                     $[\because e=\frac{1}{2}]$

Now, $b^2=a^2(1–e^2)$

$=16(1–\frac{1}{4})=12 \Rightarrow b=2\sqrt{3}$

Since, the circle of minimum area enclosing the ellipse has a radius equal to semi-major axis i.e., a

$\therefore$   Radius = a = 4

Height of $△PQR=\mathrm{radius}+2\sqrt{3}=4+2\sqrt{3}$

Hence, required area

          $=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{2}\times 2a\times (4+2\sqrt{3})$

          $=4(4+2\sqrt{3})=8(2+\sqrt{3}) \mathrm{sq}. \mathrm{units}$.

(3)

We have, two foci (2, 5), (2, –3) and eccentricity =4/5

$\Rightarrow 2be=8 \Rightarrow be=4 \Rightarrow b\left(\frac{4}{5}\right)=4 \Rightarrow b=5$

$\therefore c^2=b^2–a^2 \Rightarrow 16=25–a^2 \Rightarrow a = 3$

$\therefore$   Length of latus rectum = $\frac{2a^2}{b}=\frac{18}{5}$

(2)

We have, the ellipse $\left(E\right) : \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, (a > b)

$\Rightarrow x^2+\frac{a^2{y}^2}{b^2}=a^2$          ... (i)

The circle is $x^2+y^2=a^2{e}^2$         ... (ii)

Using (i) and (ii),

$\Rightarrow y^2(1–\frac{a^2}{b^2})=a^2(e^2–1)=a^2(1–\frac{b^2}{a^2}–1)=–b^2$

$\Rightarrow y^2\frac{(b^2–a^2)}{b^2}=–b^2 \Rightarrow y^2\frac{b^4}{a^2–b^2} \Rightarrow \left|y\right|=\frac{b^2}{\sqrt{a^2–b^2}}$

Area of triangle $=\frac{1}{2}\times 2ae\times \frac{b^2}{\sqrt{a^2–b^2}}=30$

$\Rightarrow \frac{a{b}^{2}e}{a\sqrt{1–{\displaystyle \frac{b^2}{a^2}}}}=b^2=30$

Given, $2a=17 \Rightarrow a=\frac{17}{2}$.

$\therefore$   Distance between the foci = 2ae

$=17\sqrt{1–\frac{b^2}{a^2}}=17\sqrt{1–\frac{30\times 4}{289}}=13$.

(1)

We have ellipse $E : \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$

and line y = x + p $\Rightarrow$ slope, m = 1

E and line y = x + p has point of contacts as A and B.

So, the point of contact

$=(\frac{\mp a^{2}m}{\sqrt{a^2{m}^2+b^2}},\frac{\pm b^2}{\sqrt{a^2{m}^2+b^2}})=(\frac{\mp 16}{5},\frac{\pm 9}{5})$

Then,  $A(\frac{–16}{5},\frac{9}{5}) \mathrm{and} B(\frac{16}{5},\frac{–9}{5})$

Now, line y = x intersects with ellipse E at

            $D(\frac{–12}{5},\frac{–12}{5}) \mathrm{and} C(\frac{12}{5},\frac{12}{5})$

ABCD does not form any quadrilateral but if we do not consider the order then we have,

Area of $△ABC=\frac{1}{2}$$\left|\begin{array}{ccc}–\frac{16}{5}& \frac{9}{5}& 1\\ \frac{16}{5}& –\frac{9}{5}& 1\\ \frac{12}{5}& \frac{12}{5}& 1\end{array}\right|$$=12$

$\therefore$   Area of quadrilateral ABCD = 2 (Area of $△$ABC) = 24 sq. units

(1)

For infinitely many solutions, we have, D = 0

$\Rightarrow$$\left|\begin{array}{ccc}1& 5& –1\\ 4& 3& –3\\ 24& 1& \lambda \end{array}\right|$$=0$

$\Rightarrow 1(3\lambda +3)–5(4\lambda +72)–1(4–72)=0$

$\Rightarrow 17\lambda =–289 \Rightarrow \lambda =–17$          ... (i)

Now, $D_1=0$

$\Rightarrow$$\left|\begin{array}{ccc}1& 5& –1\\ 7& 3& –3\\ \mu & 1& –17\end{array}\right|$$=0$

$\Rightarrow 1(–51+3)–5(–119+3\mu )–1(7–3\mu )=0$

$\Rightarrow –48+595–15\mu –7+3\mu =0 \Rightarrow 12\mu =540$

$\Rightarrow \mu =45$          ... (ii)

Using (i) and (ii), we get

$x+5y–z=1, 4x+3y–3z=7, 24x+y–17z=45$

$\Rightarrow z=x+5y–1$

$\therefore 4x+3y–3x–15y+3=7$

$\Rightarrow x–12y=4$

$\Rightarrow x=4+12y \mathrm{and} z=4+12y+5y–1=3+17y$

$\therefore$   (x, y, z) = (4 + 12k, k, 3 + 17k)          ($\because$ Assume y = k)

Also, $7\le 7+30k\le 77$

$\Rightarrow 0\le 30k<70$

$\Rightarrow 0\le k\le 2.3 \Rightarrow k=0,1,2$

Thus, there are three possible solutions.

(3)

Total triangles $={}^{n}C_3$

Total quadrilaterals $={}^{n}C_4$

$\because {}^{n}C_3+{}^{n}C_4=126 \Rightarrow {}^{n+1}C_4=126$

$\Rightarrow n=8$

The ellipse is $\frac{x^2}{16}+\frac{y^2}{n}=1 \Rightarrow \frac{x^2}{16}+\frac{y^2}{8}=1$

$\therefore e=\sqrt{1–\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}}$

(3)

Given, equation of circle is $x^2+y^2–2x–4y–11=0$

It can be written as

          ${(x–1)}^2+{(y–2)}^2=16$

$\therefore$   Centre of circle is (1, 2) and radius is 4.

Now, ellipse $3x^2+p{y}^2=4$ passes through (1, 2)

$\Rightarrow 3+4p=4 \Rightarrow p=\frac{1}{4}$

$\therefore$   Given equation of ellipse is $3x^2+\frac{1}{4}{y}^2=4$

i.e., $\frac{x^2}{4/3}+\frac{y^2}{16}=1$

$\therefore a^2=\frac{4}{3} \mathrm{and} b^2=16 \Rightarrow a=\frac{2}{\sqrt{3}} \mathrm{and} b=4$

Now, eccentricity of ellipse $=\sqrt{\frac{1–a^2}{b^2}}=\sqrt{1–\frac{{\displaystyle \frac{4}{3}}}{16}}=\sqrt{\frac{11}{12}}$

Now, focal distance of ellipse from (1, 2) $=4\pm \sqrt{\frac{11}{12}}\times 2$

$\therefore f_1=4+\frac{2\sqrt{11}}{2\sqrt{3}}=4+\sqrt{\frac{11}{3}} \mathrm{and} f_2=4–\sqrt{\frac{11}{3}}$

Now, $f_1{f}_2=16–\frac{11}{3}=\frac{37}{3}$

$\therefore 6f_1{f}_2–r=74–4=70$.

(3)

We have, $\frac{x^2}{4}+\frac{y^2}{2}=1$          ,,, (i)

Mid-point of chord is $(1,\frac{1}{2})$

The equation of chord to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ bisected at the point $(x_1,y_1)$ is given by

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}–1=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}–1$

$\therefore \frac{x·1}{4}+\frac{y·1/2}{2}–1=\frac{1}{4}+\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2}{2}–1$

$\Rightarrow x+y=3/2$          ... (ii)

On solving equation (i) and (ii), we get

$x=\frac{6\pm \sqrt{30}}{6} \mathrm{and} y=\frac{3}{2}–\left(\frac{6\pm \sqrt{30}}{6}\right)$

Let $x_1=\frac{6+\sqrt{30}}{6}, x_2=\frac{6–\sqrt{30}}{6}$

and $y_1=\frac{3}{2}–\left(\frac{6+\sqrt{30}}{6}\right), y_2=\frac{3}{2}–\left(\frac{6–\sqrt{30}}{6}\right)$

$\therefore$   The length of chord

$=\sqrt{(\frac{6+\sqrt{30}}{6}–{\left(\frac{6–\sqrt{30}}{6}\right))}^2+(\frac{3}{2}–\left(\frac{6+\sqrt{30}}{6}\right)–\frac{3}{2}+{\left(\frac{6–\sqrt{30}}{6}\right))}^2}$

$=\sqrt{\frac{30}{9}+\frac{30}{9}}=\sqrt{\frac{60}{9}}$

$=\frac{2}{3}\sqrt{15}$