Class 11: Ellipse- Exercise 26.1

$\displaystyle \textbf{Question 1: }\text{Find the equation of the ellipse whose focus is }(1,-2),\text{ the directrix}$
$\displaystyle 3x-2y+5=0\text{ and eccentricity is equal to }\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the ellipse and let }S(1,-2)\text{ be its focus.}$
$\displaystyle \text{Let }PM\text{ be the perpendicular distance of }P\text{ from the directrix.}$
$\displaystyle \text{By the focus-directrix definition of an ellipse,}$
$\displaystyle SP=e\cdot PM$
$\displaystyle \Rightarrow SP=\frac{1}{2}PM$
$\displaystyle \Rightarrow 4SP^2=PM^2$
$\displaystyle SP^2=(x-1)^2+(y+2)^2$
$\displaystyle PM=\frac{|3x-2y+5|}{\sqrt{3^2+(-2)^2}}=\frac{|3x-2y+5|}{\sqrt{13}}$
$\displaystyle \therefore 4\left[(x-1)^2+(y+2)^2\right]=\frac{(3x-2y+5)^2}{13}$
$\displaystyle \Rightarrow 52\left[(x-1)^2+(y+2)^2\right]=(3x-2y+5)^2$
$\displaystyle \Rightarrow 52(x^2+y^2-2x+4y+5)=(3x-2y+5)^2$
$\displaystyle \Rightarrow 52x^2+52y^2-104x+208y+260$
$\displaystyle \hspace{1.5cm}=9x^2+4y^2-12xy+30x-20y+25$
$\displaystyle \Rightarrow 43x^2+48y^2+12xy-134x+228y+235=0$
$\displaystyle \therefore \text{The required equation of the ellipse is}$
$\displaystyle 43x^2+48y^2+12xy-134x+228y+235=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the equation of the ellipse in each of the following cases:}$
$\displaystyle \text{(i) Focus is }(0,1),\text{ directrix is }x+y=0\text{ and }e=\frac{1}{2}.$
$\displaystyle \text{(ii) Focus is }(-1,1),\text{ directrix is }x-y+3=0\text{ and }e=\frac{1}{2}.$
$\displaystyle \text{(iii) Focus is }(-2,3),\text{ directrix is }2x+3y+4=0\text{ and }e=\frac{4}{5}.$
$\displaystyle \text{(iv) Focus is }(1,2),\text{ directrix is }3x+4y-5=0\text{ and }e=\frac{1}{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Let }P(x,y)\text{ be any point on the ellipse and }S(0,1)\text{ be its focus.}$
$\displaystyle \text{Let }PM\text{ be the perpendicular distance of }P\text{ from the directrix.}$
$\displaystyle \text{By the focus-directrix definition of an ellipse,}$
$\displaystyle SP=e\cdot PM$
$\displaystyle \Rightarrow SP=\frac{1}{2}PM$
$\displaystyle \Rightarrow 4SP^2=PM^2$
$\displaystyle 4\left[x^2+(y-1)^2\right]=\left[\frac{x+y}{\sqrt{1^2+1^2}}\right]^2$
$\displaystyle \Rightarrow 4(x^2+y^2-2y+1)=\frac{(x+y)^2}{2}$
$\displaystyle \Rightarrow 8(x^2+y^2-2y+1)=x^2+y^2+2xy$
$\displaystyle \Rightarrow 7x^2+7y^2-2xy-16y+8=0$
$\displaystyle \therefore \text{The required equation is }7x^2+7y^2-2xy-16y+8=0.$

$\displaystyle \text{(ii) Let }P(x,y)\text{ be any point on the ellipse and }S(-1,1)\text{ be its focus.}$
$\displaystyle \text{By the focus-directrix definition of an ellipse,}$
$\displaystyle SP=e\cdot PM$
$\displaystyle \Rightarrow SP=\frac{1}{2}PM$
$\displaystyle \Rightarrow 4SP^2=PM^2$
$\displaystyle 4\left[(x+1)^2+(y-1)^2\right]=\left[\frac{x-y+3}{\sqrt{1^2+(-1)^2}}\right]^2$
$\displaystyle \Rightarrow 8(x^2+y^2+2x-2y+2)=(x-y+3)^2$
$\displaystyle \Rightarrow 8(x^2+y^2+2x-2y+2)$
$\displaystyle \hspace{1.5cm}=x^2+y^2-2xy+6x-6y+9$
$\displaystyle \Rightarrow 7x^2+7y^2+2xy+10x-10y+7=0$
$\displaystyle \therefore \text{The required equation is }7x^2+7y^2+2xy+10x-10y+7=0.$

$\displaystyle \text{(iii) Let }P(x,y)\text{ be any point on the ellipse and }S(-2,3)\text{ be its focus.}$
$\displaystyle \text{By the focus-directrix definition of an ellipse,}$
$\displaystyle SP=e\cdot PM$
$\displaystyle \Rightarrow SP=\frac{4}{5}PM$
$\displaystyle \Rightarrow 25SP^2=16PM^2$
$\displaystyle 25\left[(x+2)^2+(y-3)^2\right]$
$\displaystyle \hspace{1.5cm}=16\left[\frac{2x+3y+4}{\sqrt{2^2+3^2}}\right]^2$
$\displaystyle \Rightarrow 325(x^2+y^2+4x-6y+13)=16(2x+3y+4)^2$
$\displaystyle \Rightarrow 325x^2+325y^2+1300x-1950y+4225$
$\displaystyle \hspace{1.5cm}=64x^2+144y^2+192xy+256x+384y+256$
$\displaystyle \Rightarrow 261x^2+181y^2-192xy+1044x-2334y+3969=0$
$\displaystyle \therefore \text{The required equation is}$
$\displaystyle 261x^2+181y^2-192xy+1044x-2334y+3969=0.$

$\displaystyle \text{(iv) Let }P(x,y)\text{ be any point on the ellipse and }S(1,2)\text{ be its focus.}$
$\displaystyle \text{By the focus-directrix definition of an ellipse,}$
$\displaystyle SP=e\cdot PM$
$\displaystyle \Rightarrow SP=\frac{1}{2}PM$
$\displaystyle \Rightarrow 4SP^2=PM^2$
$\displaystyle 4\left[(x-1)^2+(y-2)^2\right]=\left[\frac{3x+4y-5}{\sqrt{3^2+4^2}}\right]^2$
$\displaystyle \Rightarrow 100(x^2+y^2-2x-4y+5)=(3x+4y-5)^2$
$\displaystyle \Rightarrow 100x^2+100y^2-200x-400y+500$
$\displaystyle \hspace{1.5cm}=9x^2+16y^2+24xy-30x-40y+25$
$\displaystyle \Rightarrow 91x^2+84y^2-24xy-170x-360y+475=0$
$\displaystyle \therefore \text{The required equation is}$
$\displaystyle 91x^2+84y^2-24xy-170x-360y+475=0.$
$\displaystyle \\$

$\\$

$\displaystyle \textbf{Question 3: }\text{Find the eccentricity, coordinates of the foci and length of the}$
$\displaystyle \text{latus rectum of each of the following ellipses:}$
$\displaystyle \text{(i) }4x^2+9y^2=1\qquad \text{(ii) }5x^2+4y^2=1$
$\displaystyle \text{(iii) }4x^2+3y^2=1\qquad \text{(iv) }25x^2+16y^2=1600$
$\displaystyle \text{(v) }9x^2+25y^2=225$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Given }4x^2+9y^2=1$
$\displaystyle \Rightarrow \frac{x^2}{\frac14}+\frac{y^2}{\frac19}=1$
$\displaystyle \text{Comparing with }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\text{ we get}$
$\displaystyle a^2=\frac14,\qquad b^2=\frac19,\qquad a=\frac12,\qquad b=\frac13$
$\displaystyle \text{Since }a>b,\text{ the major axis lies along the }x\text{-axis.}$
$\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{\frac19}{\frac14}}=\sqrt{1-\frac49}=\frac{\sqrt5}{3}$
$\displaystyle ae=\frac12\cdot\frac{\sqrt5}{3}=\frac{\sqrt5}{6}$
$\displaystyle \therefore \text{The coordinates of the foci are }\left(\pm\frac{\sqrt5}{6},0\right).$
$\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}$
$\displaystyle =\frac{2\left(\frac19\right)}{\frac12}=\frac49$

$\displaystyle \text{(ii) Given }5x^2+4y^2=1$
$\displaystyle \Rightarrow \frac{x^2}{\frac15}+\frac{y^2}{\frac14}=1$
$\displaystyle a^2=\frac15,\qquad b^2=\frac14,\qquad a=\frac{1}{\sqrt5},\qquad b=\frac12$
$\displaystyle \text{Since }b>a,\text{ the major axis lies along the }y\text{-axis.}$
$\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}$
$\displaystyle =\sqrt{1-\frac{\frac15}{\frac14}}=\sqrt{1-\frac45}=\frac{1}{\sqrt5}$
$\displaystyle be=\frac12\cdot\frac{1}{\sqrt5}=\frac{1}{2\sqrt5}$
$\displaystyle \therefore \text{The coordinates of the foci are }\left(0,\pm\frac{1}{2\sqrt5}\right).$
$\displaystyle \text{Length of the latus rectum}=\frac{2a^2}{b}$
$\displaystyle =\frac{2\left(\frac15\right)}{\frac12}=\frac45$

$\displaystyle \text{(iii) Given }4x^2+3y^2=1$
$\displaystyle \Rightarrow \frac{x^2}{\frac14}+\frac{y^2}{\frac13}=1$
$\displaystyle a^2=\frac14,\qquad b^2=\frac13,\qquad a=\frac12,\qquad b=\frac{1}{\sqrt3}$
$\displaystyle \text{Since }b>a,\text{ the major axis lies along the }y\text{-axis.}$
$\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}$
$\displaystyle =\sqrt{1-\frac{\frac14}{\frac13}}=\sqrt{1-\frac34}=\frac12$
$\displaystyle be=\frac{1}{\sqrt3}\cdot\frac12=\frac{1}{2\sqrt3}$
$\displaystyle \therefore \text{The coordinates of the foci are }\left(0,\pm\frac{1}{2\sqrt3}\right).$
$\displaystyle \text{Length of the latus rectum}=\frac{2a^2}{b}$
$\displaystyle =\frac{2\left(\frac14\right)}{\frac{1}{\sqrt3}}=\frac{\sqrt3}{2}$

$\displaystyle \text{(iv) Given }25x^2+16y^2=1600$
$\displaystyle \Rightarrow \frac{x^2}{64}+\frac{y^2}{100}=1$
$\displaystyle a^2=64,\qquad b^2=100,\qquad a=8,\qquad b=10$
$\displaystyle \text{Since }b>a,\text{ the major axis lies along the }y\text{-axis.}$
$\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}$
$\displaystyle =\sqrt{1-\frac{64}{100}}=\sqrt{\frac{36}{100}}=\frac35$
$\displaystyle be=10\cdot\frac35=6$
$\displaystyle \therefore \text{The coordinates of the foci are }(0,\pm6).$
$\displaystyle \text{Length of the latus rectum}=\frac{2a^2}{b}$
$\displaystyle =\frac{2(64)}{10}=\frac{64}{5}$

$\displaystyle \text{(v) Given }9x^2+25y^2=225$
$\displaystyle \Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$
$\displaystyle a^2=25,\qquad b^2=9,\qquad a=5,\qquad b=3$
$\displaystyle \text{Since }a>b,\text{ the major axis lies along the }x\text{-axis.}$
$\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac45$
$\displaystyle ae=5\cdot\frac45=4$
$\displaystyle \therefore \text{The coordinates of the foci are }(\pm4,0).$
$\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}$
$\displaystyle =\frac{2(9)}{5}=\frac{18}{5}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the equation of the ellipse, referred to its major and minor axes as the}$
$\displaystyle x\text{-axis and }y\text{-axis respectively, which passes through }(-3,1)\text{ and has}$
$\displaystyle \text{eccentricity }\sqrt{\frac{2}{5}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{We know that }e=\sqrt{1-\frac{b^2}{a^2}}.$
$\displaystyle \text{Given }e=\sqrt{\frac{2}{5}}.$
$\displaystyle \therefore \sqrt{\frac{2}{5}}=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle \Rightarrow \frac{2}{5}=1-\frac{b^2}{a^2}$
$\displaystyle \Rightarrow \frac{b^2}{a^2}=\frac{3}{5}$
$\displaystyle \Rightarrow b^2=\frac{3a^2}{5}. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Since the ellipse passes through }(-3,1),\text{ substituting it in (i), we get}$
$\displaystyle \frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1$
$\displaystyle \Rightarrow \frac{9}{a^2}+\frac{1}{\frac{3a^2}{5}}=1$
$\displaystyle \Rightarrow \frac{9}{a^2}+\frac{5}{3a^2}=1$
$\displaystyle \Rightarrow \frac{1}{a^2}\left(9+\frac{5}{3}\right)=1$
$\displaystyle \Rightarrow \frac{32}{3a^2}=1$
$\displaystyle \Rightarrow a^2=\frac{32}{3}. \qquad \text{... ... ... ... ... (iii)}$
$\displaystyle \text{Substituting }a^2=\frac{32}{3}\text{ in (ii), we get}$
$\displaystyle b^2=\frac{3}{5}\times\frac{32}{3}=\frac{32}{5}.$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1$
$\displaystyle \Rightarrow \frac{3x^2}{32}+\frac{5y^2}{32}=1$
$\displaystyle \Rightarrow 3x^2+5y^2=32.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the equation of the ellipse in each of the following cases:}$
$\displaystyle \text{(i) Eccentricity }e=\frac{1}{2}\text{ and foci }(\pm2,0).$
$\displaystyle \text{(ii) Eccentricity }e=\frac{2}{3}\text{ and length of the latus rectum is }5.$
$\displaystyle \text{(iii) Eccentricity }e=\frac{1}{2}\text{ and semi-major axis is }4.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Given }e=\frac{1}{2}\text{ and the foci are }(\pm2,0).$
$\displaystyle \text{For the ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\text{ the foci are }(\pm ae,0).$
$\displaystyle \therefore ae=2$
$\displaystyle \Rightarrow a\left(\frac{1}{2}\right)=2$
$\displaystyle \Rightarrow a=4$
$\displaystyle \therefore a^2=16$
$\displaystyle \text{Also, }e^2=1-\frac{b^2}{a^2}.$
$\displaystyle \Rightarrow \frac{1}{4}=1-\frac{b^2}{16}$
$\displaystyle \Rightarrow \frac{b^2}{16}=\frac{3}{4}$
$\displaystyle \Rightarrow b^2=12$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{16}+\frac{y^2}{12}=1$
$\displaystyle \Rightarrow 3x^2+4y^2=48.$

$\displaystyle \text{(ii) Given }e=\frac{2}{3}\text{ and length of the latus rectum }=5.$
$\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}$
$\displaystyle \therefore \frac{2b^2}{a}=5$
$\displaystyle \Rightarrow b^2=\frac{5a}{2}. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{Also, }e^2=1-\frac{b^2}{a^2}.$
$\displaystyle \Rightarrow \frac{4}{9}=1-\frac{b^2}{a^2}$
$\displaystyle \Rightarrow \frac{b^2}{a^2}=\frac{5}{9}$
$\displaystyle \Rightarrow b^2=\frac{5a^2}{9}. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle \frac{5a}{2}=\frac{5a^2}{9}$
$\displaystyle \Rightarrow 9a=2a^2$
$\displaystyle \Rightarrow a=\frac{9}{2}$
$\displaystyle \therefore a^2=\frac{81}{4}$
$\displaystyle b^2=\frac{5}{2}\times\frac{9}{2}=\frac{45}{4}$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1$
$\displaystyle \Rightarrow \frac{4x^2}{81}+\frac{4y^2}{45}=1$
$\displaystyle \Rightarrow 20x^2+36y^2=405.$

$\displaystyle \text{(iii) Given }e=\frac{1}{2}\text{ and semi-major axis }a=4.$
$\displaystyle \therefore a^2=16$
$\displaystyle \text{Also, }e^2=1-\frac{b^2}{a^2}.$
$\displaystyle \Rightarrow \frac{1}{4}=1-\frac{b^2}{16}$
$\displaystyle \Rightarrow \frac{b^2}{16}=\frac{3}{4}$
$\displaystyle \Rightarrow b^2=12$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{16}+\frac{y^2}{12}=1$
$\displaystyle \Rightarrow 3x^2+4y^2=48.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the equation of the ellipse in each of the following cases:}$
$\displaystyle \text{(iv) Eccentricity }e=\frac{1}{2}\text{ and length of the major axis is }12.$
$\displaystyle \text{(v) The ellipse passes through }(1,4)\text{ and }(-6,1).$
$\displaystyle \text{(vi) Vertices are }(\pm5,0)\text{ and foci are }(\pm4,0).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(iv) Given }e=\frac{1}{2}\text{ and length of the major axis }=12.$
$\displaystyle \therefore 2a=12$
$\displaystyle \Rightarrow a=6$
$\displaystyle \therefore a^2=36$
$\displaystyle \text{We know that }e^2=1-\frac{b^2}{a^2}.$
$\displaystyle \Rightarrow \frac{1}{4}=1-\frac{b^2}{36}$
$\displaystyle \Rightarrow \frac{b^2}{36}=\frac{3}{4}$
$\displaystyle \Rightarrow b^2=27$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{36}+\frac{y^2}{27}=1$
$\displaystyle \Rightarrow 3x^2+4y^2=108.$

$\displaystyle \text{(v) Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{Since the ellipse passes through }(1,4),$
$\displaystyle \frac{1}{a^2}+\frac{16}{b^2}=1. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Let }\frac{1}{a^2}=\alpha\text{ and }\frac{1}{b^2}=\beta.$
$\displaystyle \therefore \alpha+16\beta=1. \qquad \text{... ... ... ... ... (iii)}$
$\displaystyle \text{Since the ellipse also passes through }(-6,1),$
$\displaystyle \frac{36}{a^2}+\frac{1}{b^2}=1$
$\displaystyle \therefore 36\alpha+\beta=1. \qquad \text{... ... ... ... ... (iv)}$
$\displaystyle \text{Multiplying (iii) by }36,\text{ we get}$
$\displaystyle 36\alpha+576\beta=36.$
$\displaystyle \text{Subtracting (iv),}$
$\displaystyle 575\beta=35$
$\displaystyle \Rightarrow \beta=\frac{7}{115}$
$\displaystyle \text{Substituting }\beta=\frac{7}{115}\text{ in (iii), we get}$
$\displaystyle \alpha+16\left(\frac{7}{115}\right)=1$
$\displaystyle \Rightarrow \alpha=\frac{3}{115}$
$\displaystyle \therefore \frac{1}{a^2}=\frac{3}{115}\text{ and }\frac{1}{b^2}=\frac{7}{115}.$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{3x^2}{115}+\frac{7y^2}{115}=1$
$\displaystyle \Rightarrow 3x^2+7y^2=115.$

$\displaystyle \text{(vi) Given vertices }(\pm5,0)\text{ and foci }(\pm4,0).$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$
$\displaystyle \text{Its vertices and foci are }(\pm a,0)\text{ and }(\pm ae,0)\text{ respectively.}$
$\displaystyle \therefore a=5\text{ and }ae=4$
$\displaystyle \Rightarrow e=\frac{4}{5}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=25\left(1-\frac{16}{25}\right)=9$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{25}+\frac{y^2}{9}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the equation of the ellipse in each of the following cases:}$
$\displaystyle \text{(vii) Vertices are }(0,\pm13)\text{ and foci are }(0,\pm5).$
$\displaystyle \text{(viii) Vertices are }(\pm6,0)\text{ and foci are }(\pm4,0).$
$\displaystyle \text{(ix) Ends of the major axis are }(\pm3,0)\text{ and ends of the minor axis are}$
$\displaystyle (0,\pm2).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(vii) Given vertices }(0,\pm13)\text{ and foci }(0,\pm5).$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad b>a.$
$\displaystyle \text{Its vertices and foci are }(0,\pm b)\text{ and }(0,\pm be)\text{ respectively.}$
$\displaystyle \therefore b=13\text{ and }be=5$
$\displaystyle \Rightarrow e=\frac{5}{13}$
$\displaystyle \text{Also, }a^2=b^2(1-e^2).$
$\displaystyle \therefore a^2=169\left(1-\frac{25}{169}\right)=144$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{144}+\frac{y^2}{169}=1.$

$\displaystyle \text{(viii) Given vertices }(\pm6,0)\text{ and foci }(\pm4,0).$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b.$
$\displaystyle \text{Its vertices and foci are }(\pm a,0)\text{ and }(\pm ae,0)\text{ respectively.}$
$\displaystyle \therefore a=6\text{ and }ae=4$
$\displaystyle \Rightarrow e=\frac{4}{6}=\frac{2}{3}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=36\left(1-\frac{4}{9}\right)=36\left(\frac{5}{9}\right)=20$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{36}+\frac{y^2}{20}=1.$

$\displaystyle \text{(ix) Given ends of the major axis }(\pm3,0)\text{ and ends of the minor axis }(0,\pm2).$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$
$\displaystyle \text{The endpoints of its major and minor axes are }(\pm a,0)\text{ and }(0,\pm b).$
$\displaystyle \therefore a=3\text{ and }b=2$
$\displaystyle \therefore a^2=9\text{ and }b^2=4$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the equation of the ellipse in each of the following cases:}$
$\displaystyle \text{(x) Ends of the major axis are }(0,\pm\sqrt5)\text{ and ends of the minor axis are }(\pm1,0).$
$\displaystyle \text{(xi) Length of the major axis is }26\text{ and the foci are }(\pm5,0).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(x) Given ends of the major axis }(0,\pm\sqrt5)\text{ and ends of the minor axis }(\pm1,0).$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad b>a.$
$\displaystyle \text{The endpoints of its major and minor axes are }(0,\pm b)\text{ and }(\pm a,0).$
$\displaystyle \therefore b=\sqrt5\text{ and }a=1$
$\displaystyle \therefore b^2=5\text{ and }a^2=1$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{1}+\frac{y^2}{5}=1$
$\displaystyle \Rightarrow x^2+\frac{y^2}{5}=1.$

$\displaystyle \text{(xi) Given length of the major axis }=26\text{ and foci }(\pm5,0).$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b.$
$\displaystyle \text{Since the length of the major axis is }26,$
$\displaystyle 2a=26$
$\displaystyle \Rightarrow a=13$
$\displaystyle \therefore a^2=169$
$\displaystyle \text{The foci are }(\pm ae,0).$
$\displaystyle \therefore ae=5$
$\displaystyle \Rightarrow e=\frac{5}{13}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=169\left(1-\frac{25}{169}\right)=144$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{169}+\frac{y^2}{144}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the equation of the ellipse in each of the following cases:}$
$\displaystyle \text{(xii) Length of the minor axis is }16\text{ and the foci are }(0,\pm6).$
$\displaystyle \text{(xiii) Foci are }(\pm3,0)\text{ and }a=4.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(xii) Given length of the minor axis }=16\text{ and foci }(0,\pm6).$
$\displaystyle \text{Since the foci lie on the }y\text{-axis, the major axis lies along the }y\text{-axis.}$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad b>a.$
$\displaystyle \text{The length of the minor axis is }2a.$
$\displaystyle \therefore 2a=16$
$\displaystyle \Rightarrow a=8$
$\displaystyle \therefore a^2=64$
$\displaystyle \text{The foci are }(0,\pm be).$
$\displaystyle \therefore be=6$
$\displaystyle \text{Also, }a^2=b^2(1-e^2)=b^2-b^2e^2.$
$\displaystyle \therefore a^2=b^2-(be)^2$
$\displaystyle \Rightarrow 64=b^2-36$
$\displaystyle \Rightarrow b^2=100$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{64}+\frac{y^2}{100}=1.$

$\displaystyle \text{(xiii) Given foci }(\pm3,0)\text{ and semi-major axis }a=4.$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b.$
$\displaystyle \text{The foci are }(\pm ae,0).$
$\displaystyle \therefore ae=3$
$\displaystyle \Rightarrow 4e=3$
$\displaystyle \Rightarrow e=\frac{3}{4}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=16\left(1-\frac{9}{16}\right)=7$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{16}+\frac{y^2}{7}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Find the equation of the ellipse whose foci are }(4,0)\text{ and }(-4,0),$
$\displaystyle \text{and whose eccentricity is }\frac{1}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{The coordinates of its foci are }(\pm ae,0).$
$\displaystyle \text{Given }e=\frac{1}{3}\text{ and the foci are }(\pm4,0).$
$\displaystyle \therefore ae=4$
$\displaystyle \Rightarrow a\left(\frac{1}{3}\right)=4$
$\displaystyle \Rightarrow a=12$
$\displaystyle \therefore a^2=144$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=12^2\left[1-\left(\frac{1}{3}\right)^2\right]$
$\displaystyle =144\left(1-\frac{1}{9}\right)$
$\displaystyle =144\left(\frac{8}{9}\right)=128$
$\displaystyle \text{Substituting }a^2=144\text{ and }b^2=128\text{ in (i), we get}$
$\displaystyle \frac{x^2}{144}+\frac{y^2}{128}=1.$
$\displaystyle \therefore \text{The required equation of the ellipse is }\frac{x^2}{144}+\frac{y^2}{128}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the equation of the ellipse in standard form whose minor axis is equal}$
$\displaystyle \text{to the distance between its foci and whose latus rectum is }10.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b.$
$\displaystyle \text{Length of the minor axis}=2b$
$\displaystyle \text{Distance between the foci}=2ae$
$\displaystyle \text{Since the minor axis is equal to the distance between the foci,}$
$\displaystyle 2b=2ae$
$\displaystyle \Rightarrow b=ae$
$\displaystyle \Rightarrow b^2=a^2e^2$
$\displaystyle \text{We know that }e^2=1-\frac{b^2}{a^2}.$
$\displaystyle \therefore b^2=a^2\left(1-\frac{b^2}{a^2}\right)$
$\displaystyle \Rightarrow b^2=a^2-b^2$
$\displaystyle \Rightarrow a^2=2b^2. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{Length of the latus rectum}=\frac{2b^2}{a}$
$\displaystyle \therefore \frac{2b^2}{a}=10$
$\displaystyle \Rightarrow b^2=5a. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Substituting }b^2=5a\text{ in (i), we get}$
$\displaystyle a^2=2(5a)$
$\displaystyle \Rightarrow a^2=10a$
$\displaystyle \Rightarrow a=10$
$\displaystyle \therefore a^2=100$
$\displaystyle \text{From (ii),}$
$\displaystyle b^2=5(10)=50$
$\displaystyle \therefore \text{The equation of the ellipse is}$
$\displaystyle \frac{x^2}{100}+\frac{y^2}{50}=1$
$\displaystyle \Rightarrow x^2+2y^2=100.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find the equation of the ellipse whose center is }(-2,3)\text{ and whose}$
$\displaystyle \text{semi-axes are }3\text{ and }2\text{ when the major axis is:}$
$\displaystyle \text{(i) parallel to the }x\text{-axis}\qquad\text{(ii) parallel to the }y\text{-axis}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The center of the ellipse is }(h,k)=(-2,3).$
$\displaystyle \text{The semi-major axis is }3\text{ and the semi-minor axis is }2.$

$\displaystyle \text{(i) When the major axis is parallel to the }x\text{-axis,}$
$\displaystyle \text{the equation of the ellipse is}$
$\displaystyle \frac{(x-h)^2}{3^2}+\frac{(y-k)^2}{2^2}=1.$
$\displaystyle \therefore \frac{(x+2)^2}{9}+\frac{(y-3)^2}{4}=1$
$\displaystyle \Rightarrow 4(x+2)^2+9(y-3)^2=36$
$\displaystyle \Rightarrow 4(x^2+4x+4)+9(y^2-6y+9)=36$
$\displaystyle \Rightarrow 4x^2+16x+16+9y^2-54y+81=36$
$\displaystyle \Rightarrow 4x^2+9y^2+16x-54y+61=0.$
$\displaystyle \therefore \text{The required equation is }\frac{(x+2)^2}{9}+\frac{(y-3)^2}{4}=1.$

$\displaystyle \text{(ii) When the major axis is parallel to the }y\text{-axis,}$
$\displaystyle \text{the equation of the ellipse is}$
$\displaystyle \frac{(x-h)^2}{2^2}+\frac{(y-k)^2}{3^2}=1.$
$\displaystyle \therefore \frac{(x+2)^2}{4}+\frac{(y-3)^2}{9}=1$
$\displaystyle \Rightarrow 9(x+2)^2+4(y-3)^2=36$
$\displaystyle \Rightarrow 9(x^2+4x+4)+4(y^2-6y+9)=36$
$\displaystyle \Rightarrow 9x^2+36x+36+4y^2-24y+36=36$
$\displaystyle \Rightarrow 9x^2+4y^2+36x-24y+36=0.$
$\displaystyle \therefore \text{The required equation is }\frac{(x+2)^2}{4}+\frac{(y-3)^2}{9}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the eccentricity of an ellipse whose latus rectum is:}$
$\displaystyle \text{(i) half of its minor axis}\qquad\text{(ii) half of its major axis}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }2a\text{ and }2b\text{ be the lengths of the major and minor axes respectively.}$
$\displaystyle \text{(i) When the latus rectum is half of the minor axis,}$
$\displaystyle \frac{2b^2}{a}=\frac{1}{2}\times2b$
$\displaystyle \Rightarrow 2b^2=ab$
$\displaystyle \Rightarrow b=\frac{a}{2}$
$\displaystyle \Rightarrow b^2=\frac{a^2}{4}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore \frac{a^2}{4}=a^2(1-e^2)$
$\displaystyle \Rightarrow \frac{1}{4}=1-e^2$
$\displaystyle \Rightarrow e^2=\frac{3}{4}$
$\displaystyle \Rightarrow e=\frac{\sqrt3}{2}.$
$\displaystyle \text{(ii) When the latus rectum is half of the major axis,}$
$\displaystyle \frac{2b^2}{a}=\frac{1}{2}\times2a$
$\displaystyle \Rightarrow 2b^2=a^2$
$\displaystyle \Rightarrow a^2=2b^2$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=2b^2(1-e^2)$
$\displaystyle \Rightarrow 1=2-2e^2$
$\displaystyle \Rightarrow 2e^2=1$
$\displaystyle \Rightarrow e^2=\frac{1}{2}$
$\displaystyle \Rightarrow e=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Find the center, lengths of the axes, eccentricity, and foci of each ellipse:}$
$\displaystyle \text{(i) }x^2+2y^2-2x+12y+10=0$
$\displaystyle \text{(ii) }x^2+4y^2-4x+24y+31=0$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Given }x^2+2y^2-2x+12y+10=0$
$\displaystyle \Rightarrow (x^2-2x)+2(y^2+6y)=-10$
$\displaystyle \Rightarrow (x^2-2x+1)+2(y^2+6y+9)=-10+1+18$
$\displaystyle \Rightarrow (x-1)^2+2(y+3)^2=9$
$\displaystyle \Rightarrow \frac{(x-1)^2}{9}+\frac{(y+3)^2}{\frac{9}{2}}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=1,\qquad k=-3,\qquad a^2=9,\qquad b^2=\frac{9}{2}$
$\displaystyle \therefore a=3\text{ and }b=\frac{3}{\sqrt2}$
$\displaystyle \therefore \text{Center}=(1,-3)$
$\displaystyle \text{Since }a>b,\text{ the major axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Length of the major axis}=2a=2(3)=6$
$\displaystyle \text{Length of the minor axis}=2b=2\left(\frac{3}{\sqrt2}\right)=3\sqrt2$
$\displaystyle \text{Eccentricity }e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{\frac{9}{2}}{9}}$
$\displaystyle =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt2}$
$\displaystyle ae=3\left(\frac{1}{\sqrt2}\right)=\frac{3}{\sqrt2}$
$\displaystyle \therefore \text{Foci}=(h\pm ae,k)$
$\displaystyle =\left(1\pm\frac{3}{\sqrt2},-3\right)$

$\displaystyle \text{(ii) Given }x^2+4y^2-4x+24y+31=0$
$\displaystyle \Rightarrow (x^2-4x)+4(y^2+6y)=-31$
$\displaystyle \Rightarrow (x^2-4x+4)+4(y^2+6y+9)=-31+4+36$
$\displaystyle \Rightarrow (x-2)^2+4(y+3)^2=9$
$\displaystyle \Rightarrow \frac{(x-2)^2}{9}+\frac{(y+3)^2}{\frac{9}{4}}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=2,\qquad k=-3,\qquad a^2=9,\qquad b^2=\frac{9}{4}$
$\displaystyle \therefore a=3\text{ and }b=\frac{3}{2}$
$\displaystyle \therefore \text{Center}=(2,-3)$
$\displaystyle \text{Since }a>b,\text{ the major axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Length of the major axis}=2a=2(3)=6$
$\displaystyle \text{Length of the minor axis}=2b=2\left(\frac{3}{2}\right)=3$
$\displaystyle \text{Eccentricity }e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{\frac{9}{4}}{9}}$
$\displaystyle =\sqrt{\frac{3}{4}}=\frac{\sqrt3}{2}$
$\displaystyle ae=3\left(\frac{\sqrt3}{2}\right)=\frac{3\sqrt3}{2}$
$\displaystyle \therefore \text{Foci}=(h\pm ae,k)$
$\displaystyle =\left(2\pm\frac{3\sqrt3}{2},-3\right)$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Find the center, lengths of the axes, eccentricity, and foci of each ellipse:}$
$\displaystyle \text{(iii) }4x^2+y^2-8x+2y+1=0$
$\displaystyle \text{(iv) }3x^2+4y^2-12x-8y+4=0$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(iii) Given }4x^2+y^2-8x+2y+1=0$
$\displaystyle \Rightarrow 4(x^2-2x)+(y^2+2y)=-1$
$\displaystyle \Rightarrow 4(x^2-2x+1)+(y^2+2y+1)=-1+4+1$
$\displaystyle \Rightarrow 4(x-1)^2+(y+1)^2=4$
$\displaystyle \Rightarrow \frac{(x-1)^2}{1}+\frac{(y+1)^2}{4}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1,$
$\displaystyle h=1,\qquad k=-1,\qquad a^2=4,\qquad b^2=1$
$\displaystyle \therefore a=2\text{ and }b=1$
$\displaystyle \therefore \text{Center}=(1,-1)$
$\displaystyle \text{Since }a>b,\text{ the major axis is parallel to the }y\text{-axis.}$
$\displaystyle \text{Length of the major axis}=2a=2(2)=4$
$\displaystyle \text{Length of the minor axis}=2b=2(1)=2$
$\displaystyle \text{Eccentricity }e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{1}{4}}$
$\displaystyle =\sqrt{\frac{3}{4}}=\frac{\sqrt3}{2}$
$\displaystyle ae=2\left(\frac{\sqrt3}{2}\right)=\sqrt3$
$\displaystyle \therefore \text{Foci}=(h,k\pm ae)$
$\displaystyle =\left(1,-1\pm\sqrt3\right)$

$\displaystyle \text{(iv) Given }3x^2+4y^2-12x-8y+4=0$
$\displaystyle \Rightarrow 3(x^2-4x)+4(y^2-2y)=-4$
$\displaystyle \Rightarrow 3(x^2-4x+4)+4(y^2-2y+1)=-4+12+4$
$\displaystyle \Rightarrow 3(x-2)^2+4(y-1)^2=12$
$\displaystyle \Rightarrow \frac{(x-2)^2}{4}+\frac{(y-1)^2}{3}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=2,\qquad k=1,\qquad a^2=4,\qquad b^2=3$
$\displaystyle \therefore a=2\text{ and }b=\sqrt3$
$\displaystyle \therefore \text{Center}=(2,1)$
$\displaystyle \text{Since }a>b,\text{ the major axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Length of the major axis}=2a=2(2)=4$
$\displaystyle \text{Length of the minor axis}=2b=2\sqrt3$
$\displaystyle \text{Eccentricity }e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{3}{4}}$
$\displaystyle =\frac{1}{2}$
$\displaystyle ae=2\left(\frac{1}{2}\right)=1$
$\displaystyle \therefore \text{Foci}=(h\pm ae,k)$
$\displaystyle =(2\pm1,1)$
$\displaystyle \therefore \text{The foci are }(1,1)\text{ and }(3,1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Find the center, lengths of the axes, eccentricity, and foci of each ellipse:}$
$\displaystyle \text{(v) }4x^2+16y^2-24x-32y-12=0$
$\displaystyle \text{(vi) }x^2+4y^2-2x=0$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(v) Given }4x^2+16y^2-24x-32y-12=0$
$\displaystyle \Rightarrow 4(x^2-6x)+16(y^2-2y)=12$
$\displaystyle \Rightarrow 4(x^2-6x+9)+16(y^2-2y+1)=12+36+16$
$\displaystyle \Rightarrow 4(x-3)^2+16(y-1)^2=64$
$\displaystyle \Rightarrow \frac{(x-3)^2}{16}+\frac{(y-1)^2}{4}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=3,\qquad k=1,\qquad a^2=16,\qquad b^2=4$
$\displaystyle \therefore a=4\text{ and }b=2$
$\displaystyle \therefore \text{Center}=(3,1)$
$\displaystyle \text{Since }a>b,\text{ the major axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Length of the major axis}=2a=2(4)=8$
$\displaystyle \text{Length of the minor axis}=2b=2(2)=4$
$\displaystyle \text{Eccentricity }e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{4}{16}}$
$\displaystyle =\sqrt{\frac{3}{4}}=\frac{\sqrt3}{2}$
$\displaystyle ae=4\left(\frac{\sqrt3}{2}\right)=2\sqrt3$
$\displaystyle \therefore \text{Foci}=(h\pm ae,k)$
$\displaystyle =\left(3\pm2\sqrt3,1\right)$

$\displaystyle \text{(vi) Given }x^2+4y^2-2x=0$
$\displaystyle \Rightarrow (x^2-2x)+4y^2=0$
$\displaystyle \Rightarrow (x^2-2x+1)+4y^2=1$
$\displaystyle \Rightarrow (x-1)^2+4y^2=1$
$\displaystyle \Rightarrow \frac{(x-1)^2}{1}+\frac{y^2}{\frac{1}{4}}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=1,\qquad k=0,\qquad a^2=1,\qquad b^2=\frac{1}{4}$
$\displaystyle \therefore a=1\text{ and }b=\frac{1}{2}$
$\displaystyle \therefore \text{Center}=(1,0)$
$\displaystyle \text{Since }a>b,\text{ the major axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Length of the major axis}=2a=2$
$\displaystyle \text{Length of the minor axis}=2b=1$
$\displaystyle \text{Eccentricity }e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{\frac14}{1}}$
$\displaystyle =\sqrt{\frac34}=\frac{\sqrt3}{2}$
$\displaystyle ae=1\left(\frac{\sqrt3}{2}\right)=\frac{\sqrt3}{2}$
$\displaystyle \therefore \text{Foci}=(h\pm ae,k)$
$\displaystyle =\left(1\pm\frac{\sqrt3}{2},0\right)$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Find the equation of an ellipse whose foci are }(\pm3,0)\text{ and which passes}$
$\displaystyle \text{through }(4,1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{The coordinates of its foci are }(\pm ae,0).$
$\displaystyle \text{Since the foci are }(\pm3,0),$
$\displaystyle ae=3$
$\displaystyle \Rightarrow a^2e^2=9. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2)$
$\displaystyle \Rightarrow b^2=a^2-a^2e^2$
$\displaystyle \Rightarrow b^2=a^2-9. \qquad \text{... ... ... ... ... (iii)}$
$\displaystyle \text{Since the ellipse passes through }(4,1),$
$\displaystyle \frac{4^2}{a^2}+\frac{1^2}{b^2}=1$
$\displaystyle \Rightarrow \frac{16}{a^2}+\frac{1}{b^2}=1$
$\displaystyle \Rightarrow 16b^2+a^2=a^2b^2. \qquad \text{... ... ... ... ... (iv)}$
$\displaystyle \text{Substituting }b^2=a^2-9\text{ in (iv), we get}$
$\displaystyle 16(a^2-9)+a^2=a^2(a^2-9)$
$\displaystyle \Rightarrow 16a^2-144+a^2=a^4-9a^2$
$\displaystyle \Rightarrow a^4-26a^2+144=0$
$\displaystyle \Rightarrow (a^2-18)(a^2-8)=0$
$\displaystyle \Rightarrow a^2=18\qquad\text{or}\qquad a^2=8$
$\displaystyle \text{If }a^2=8,\text{ then }b^2=8-9=-1,\text{ which is impossible.}$
$\displaystyle \therefore a^2=18$
$\displaystyle \text{From (iii),}$
$\displaystyle b^2=18-9=9$
$\displaystyle \therefore \text{The required equation of the ellipse is}$
$\displaystyle \frac{x^2}{18}+\frac{y^2}{9}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Find the equation of an ellipse whose eccentricity is }\frac{2}{3},\text{ whose}$
$\displaystyle \text{latus rectum is }5,\text{ and whose center is at the origin.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Assuming that the major axis lies along the }x\text{-axis, let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{The length of the latus rectum is }\frac{2b^2}{a}.$
$\displaystyle \therefore \frac{2b^2}{a}=5$
$\displaystyle \Rightarrow b^2=\frac{5a}{2}. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \text{Given }e=\frac{2}{3}.$
$\displaystyle \therefore \frac{5a}{2}=a^2\left[1-\left(\frac{2}{3}\right)^2\right]$
$\displaystyle \Rightarrow \frac{5a}{2}=a^2\left(1-\frac{4}{9}\right)$
$\displaystyle \Rightarrow \frac{5a}{2}=\frac{5a^2}{9}$
$\displaystyle \Rightarrow \frac{1}{2}=\frac{a}{9}$
$\displaystyle \Rightarrow a=\frac{9}{2}$
$\displaystyle \therefore a^2=\frac{81}{4}$
$\displaystyle \text{Substituting }a=\frac{9}{2}\text{ in (ii), we get}$
$\displaystyle b^2=\frac{5}{2}\left(\frac{9}{2}\right)=\frac{45}{4}$
$\displaystyle \text{Substituting }a^2=\frac{81}{4}\text{ and }b^2=\frac{45}{4}\text{ in (i), we get}$
$\displaystyle \frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1$
$\displaystyle \Rightarrow \frac{4x^2}{81}+\frac{4y^2}{45}=1$
$\displaystyle \Rightarrow 20x^2+36y^2=405.$
$\displaystyle \therefore \text{The required equation of the ellipse is }\frac{4x^2}{81}+\frac{4y^2}{45}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Find the equation of an ellipse with its foci on the }y\text{-axis, eccentricity}$
$\displaystyle \frac{3}{4},\text{ center at the origin, and passing through }(6,4).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the foci lie on the }y\text{-axis, let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad b>a. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{Given }e=\frac{3}{4}.$
$\displaystyle \text{Also, }a^2=b^2(1-e^2).$
$\displaystyle \therefore a^2=b^2\left[1-\left(\frac{3}{4}\right)^2\right]$
$\displaystyle =b^2\left(1-\frac{9}{16}\right)$
$\displaystyle =\frac{7b^2}{16}. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Since the ellipse passes through }(6,4),$
$\displaystyle \frac{6^2}{a^2}+\frac{4^2}{b^2}=1$
$\displaystyle \Rightarrow \frac{36}{a^2}+\frac{16}{b^2}=1$
$\displaystyle \text{Substituting }a^2=\frac{7b^2}{16}\text{ from (ii), we get}$
$\displaystyle \frac{36}{\frac{7b^2}{16}}+\frac{16}{b^2}=1$
$\displaystyle \Rightarrow \frac{576}{7b^2}+\frac{16}{b^2}=1$
$\displaystyle \Rightarrow \frac{1}{b^2}\left(\frac{576}{7}+\frac{112}{7}\right)=1$
$\displaystyle \Rightarrow \frac{688}{7b^2}=1$
$\displaystyle \Rightarrow b^2=\frac{688}{7}$
$\displaystyle \text{Substituting }b^2=\frac{688}{7}\text{ in (ii), we get}$
$\displaystyle a^2=\frac{7}{16}\left(\frac{688}{7}\right)=43$
$\displaystyle \text{Substituting }a^2=43\text{ and }b^2=\frac{688}{7}\text{ in (i), we get}$
$\displaystyle \frac{x^2}{43}+\frac{y^2}{\frac{688}{7}}=1$
$\displaystyle \Rightarrow \frac{x^2}{43}+\frac{7y^2}{688}=1$
$\displaystyle \Rightarrow 16x^2+7y^2=688.$
$\displaystyle \therefore \text{The required equation of the ellipse is }\frac{x^2}{43}+\frac{7y^2}{688}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Find the equation of an ellipse whose axes lie along the coordinate axes}$
$\displaystyle \text{and which passes through }(4,3)\text{ and }(-1,4).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the required ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{Since the ellipse passes through }(4,3),$
$\displaystyle \frac{4^2}{a^2}+\frac{3^2}{b^2}=1$
$\displaystyle \Rightarrow \frac{16}{a^2}+\frac{9}{b^2}=1. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Since the ellipse also passes through }(-1,4),$
$\displaystyle \frac{(-1)^2}{a^2}+\frac{4^2}{b^2}=1$
$\displaystyle \Rightarrow \frac{1}{a^2}+\frac{16}{b^2}=1. \qquad \text{... ... ... ... ... (iii)}$
$\displaystyle \text{Let }\frac{1}{a^2}=u\text{ and }\frac{1}{b^2}=v.$
$\displaystyle \text{Then, equations (ii) and (iii) become}$
$\displaystyle 16u+9v=1 \qquad \text{... ... ... ... ... (iv)}$
$\displaystyle u+16v=1. \qquad \text{... ... ... ... ... (v)}$
$\displaystyle \text{Multiplying (v) by }16\text{ and subtracting (iv), we get}$
$\displaystyle 247v=15$
$\displaystyle \Rightarrow v=\frac{15}{247}$
$\displaystyle \therefore \frac{1}{b^2}=\frac{15}{247}$
$\displaystyle \Rightarrow b^2=\frac{247}{15}$
$\displaystyle \text{Substituting }v=\frac{15}{247}\text{ in (v), we get}$
$\displaystyle u+16\left(\frac{15}{247}\right)=1$
$\displaystyle \Rightarrow u=\frac{7}{247}$
$\displaystyle \therefore \frac{1}{a^2}=\frac{7}{247}$
$\displaystyle \Rightarrow a^2=\frac{247}{7}$
$\displaystyle \text{Substituting }a^2=\frac{247}{7}\text{ and }b^2=\frac{247}{15}\text{ in (i), we get}$
$\displaystyle \frac{x^2}{\frac{247}{7}}+\frac{y^2}{\frac{247}{15}}=1$
$\displaystyle \Rightarrow \frac{7x^2}{247}+\frac{15y^2}{247}=1$
$\displaystyle \Rightarrow 7x^2+15y^2=247.$
$\displaystyle \therefore \text{The required equation of the ellipse is }7x^2+15y^2=247.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Find the equation of an ellipse whose axes lie along the coordinate axes,}$
$\displaystyle \text{which passes through }(-3,1)\text{ and has eccentricity }\sqrt{\frac{2}{5}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the major axis is not specified, there are two possible cases.}$
$\displaystyle \text{Case I: The major axis lies along the }x\text{-axis.}$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{Given }e=\sqrt{\frac{2}{5}}.$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=a^2\left[1-\left(\sqrt{\frac{2}{5}}\right)^2\right]$
$\displaystyle =a^2\left(1-\frac{2}{5}\right)$
$\displaystyle =\frac{3a^2}{5}. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Since the ellipse passes through }(-3,1),$
$\displaystyle \frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1$
$\displaystyle \Rightarrow \frac{9}{a^2}+\frac{1}{b^2}=1. \qquad \text{... ... ... ... ... (iii)}$
$\displaystyle \text{Substituting }b^2=\frac{3a^2}{5}\text{ in (iii), we get}$
$\displaystyle \frac{9}{a^2}+\frac{1}{\frac{3a^2}{5}}=1$
$\displaystyle \Rightarrow \frac{1}{a^2}\left(9+\frac{5}{3}\right)=1$
$\displaystyle \Rightarrow \frac{32}{3a^2}=1$
$\displaystyle \Rightarrow a^2=\frac{32}{3}$
$\displaystyle \text{From (ii),}$
$\displaystyle b^2=\frac{3}{5}\left(\frac{32}{3}\right)=\frac{32}{5}$
$\displaystyle \therefore \frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1$
$\displaystyle \Rightarrow \frac{3x^2}{32}+\frac{5y^2}{32}=1$
$\displaystyle \Rightarrow 3x^2+5y^2=32. \qquad \text{... ... ... ... ... (iv)}$
$\displaystyle \text{Case II: The major axis lies along the }y\text{-axis.}$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1,\qquad a>b. \qquad \text{... ... ... ... ... (v)}$
$\displaystyle \text{As before, }b^2=\frac{3a^2}{5}. \qquad \text{... ... ... ... ... (vi)}$
$\displaystyle \text{Since the ellipse passes through }(-3,1),$
$\displaystyle \frac{(-3)^2}{b^2}+\frac{1^2}{a^2}=1$
$\displaystyle \Rightarrow \frac{9}{b^2}+\frac{1}{a^2}=1$
$\displaystyle \text{Substituting }b^2=\frac{3a^2}{5},\text{ we get}$
$\displaystyle \frac{9}{\frac{3a^2}{5}}+\frac{1}{a^2}=1$
$\displaystyle \Rightarrow \frac{15}{a^2}+\frac{1}{a^2}=1$
$\displaystyle \Rightarrow \frac{16}{a^2}=1$
$\displaystyle \Rightarrow a^2=16$
$\displaystyle \text{From (vi),}$
$\displaystyle b^2=\frac{3}{5}(16)=\frac{48}{5}$
$\displaystyle \therefore \frac{x^2}{\frac{48}{5}}+\frac{y^2}{16}=1$
$\displaystyle \Rightarrow \frac{5x^2}{48}+\frac{3y^2}{48}=1$
$\displaystyle \Rightarrow 5x^2+3y^2=48. \qquad \text{... ... ... ... ... (vii)}$
$\displaystyle \therefore \text{The required equations are }3x^2+5y^2=32\text{ and }5x^2+3y^2=48.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Find the equation of an ellipse if the distance between its foci is }8\text{ units}$
$\displaystyle \text{and the distance between its directrices is }18\text{ units}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the semi-major axis and }e\text{ be the eccentricity of the ellipse.}$
$\displaystyle \text{The distance between the foci is }2ae.$
$\displaystyle \therefore 2ae=8$
$\displaystyle \Rightarrow ae=4. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{The distance between the directrices is }\frac{2a}{e}.$
$\displaystyle \therefore \frac{2a}{e}=18$
$\displaystyle \Rightarrow \frac{a}{e}=9. \qquad \text{... ... ... ... ... (ii)}$
$\displaystyle \text{Multiplying (i) and (ii), we get}$
$\displaystyle (ae)\left(\frac{a}{e}\right)=4(9)$
$\displaystyle \Rightarrow a^2=36$
$\displaystyle \therefore a=6$
$\displaystyle \text{From (i),}$
$\displaystyle 6e=4$
$\displaystyle \Rightarrow e=\frac{2}{3}$
$\displaystyle \text{Also, }b^2=a^2(1-e^2).$
$\displaystyle \therefore b^2=36\left[1-\left(\frac{2}{3}\right)^2\right]$
$\displaystyle =36\left(1-\frac{4}{9}\right)$
$\displaystyle =36\left(\frac{5}{9}\right)=20$
$\displaystyle \text{If the major axis lies along the }x\text{-axis, the equation is}$
$\displaystyle \frac{x^2}{36}+\frac{y^2}{20}=1.$
$\displaystyle \text{If the major axis lies along the }y\text{-axis, the equation is}$
$\displaystyle \frac{x^2}{20}+\frac{y^2}{36}=1.$
$\displaystyle \therefore \text{The required equations are }\frac{x^2}{36}+\frac{y^2}{20}=1\text{ and }\frac{x^2}{20}+\frac{y^2}{36}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Find the equation of an ellipse whose vertices are }(0,\pm10)$
$\displaystyle \text{and whose eccentricity is }e=\frac45.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the vertices are }(0,\pm10),\text{ the major axis lies along the }y\text{-axis.}$
$\displaystyle \text{Let the equation of the ellipse be}$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad b>a. \qquad \text{... ... ... ... ... (i)}$
$\displaystyle \text{The coordinates of the vertices are }(0,\pm b)=(0,\pm10).$
$\displaystyle \therefore b=10$
$\displaystyle \Rightarrow b^2=100$
$\displaystyle \text{Also, }a^2=b^2(1-e^2).$
$\displaystyle \text{Given }e=\frac45.$
$\displaystyle \therefore a^2=100\left[1-\left(\frac45\right)^2\right]$
$\displaystyle =100\left(1-\frac{16}{25}\right)$
$\displaystyle =100\left(\frac9{25}\right)=36$
$\displaystyle \text{Substituting }a^2=36\text{ and }b^2=100\text{ in (i), we get}$
$\displaystyle \frac{x^2}{36}+\frac{y^2}{100}=1$
$\displaystyle \Rightarrow 100x^2+36y^2=3600.$
$\displaystyle \therefore \text{The required equation of the ellipse is }\frac{x^2}{36}+\frac{y^2}{100}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{A rod of length }12\text{ cm moves with its ends always touching the coordinate}$
$\displaystyle \text{axes. Determine the equation of the locus of a point }P\text{ on the rod, which is }3\text{ cm}$
$\displaystyle \text{from the end in contact with the }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }AB\text{ be the rod making an angle }\theta\text{ with }OX,\text{ and let }P\text{ be the point on it}$
$\displaystyle \text{such that }AP=3\text{ cm.}$
$\displaystyle AB=12\text{ cm}$
$\displaystyle \therefore PB=AB-AP=12-3=9\text{ cm}$
$\displaystyle \text{From }P,\text{ draw }PQ\perp OY\text{ and }PR\perp OX.$
$\displaystyle \text{In }\triangle PBQ,\text{ we have}$
$\displaystyle \cos\theta=\frac{PQ}{PB}=\frac{x}{9}$
$\displaystyle \text{In }\triangle PRA,\text{ we have}$
$\displaystyle \sin\theta=\frac{PR}{PA}=\frac{y}{3}$
$\displaystyle \text{Using }\sin^2\theta+\cos^2\theta=1,\text{ we get}$
$\displaystyle \left(\frac{y}{3}\right)^2+\left(\frac{x}{9}\right)^2=1$
$\displaystyle \Rightarrow \frac{x^2}{81}+\frac{y^2}{9}=1$
$\displaystyle \therefore \text{The required locus of }P\text{ is } \frac{x^2}{81}+\frac{y^2}{9}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Find the equation of the set of all points whose distances from }(0,4)\text{ are}$
$\displaystyle \frac{2}{3}\text{ of their distances from the line }y=9.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required locus.}$
$\displaystyle \text{Let }Q=(0,4),\text{ and let }PL\text{ be the perpendicular distance of }P\text{ from }y=9.$
$\displaystyle \text{Given }PQ=\frac{2}{3}PL.$
$\displaystyle \therefore \sqrt{(x-0)^2+(y-4)^2}=\frac{2}{3}|y-9|$
$\displaystyle \Rightarrow 9\left[x^2+(y-4)^2\right]=4(y-9)^2$
$\displaystyle \Rightarrow 9\left(x^2+y^2-8y+16\right)=4\left(y^2-18y+81\right)$
$\displaystyle \Rightarrow 9x^2+9y^2-72y+144=4y^2-72y+324$
$\displaystyle \Rightarrow 9x^2+5y^2=180$
$\displaystyle \Rightarrow \frac{x^2}{20}+\frac{y^2}{36}=1$
$\displaystyle \therefore \text{The required equation of the locus is }\frac{x^2}{20}+\frac{y^2}{36}=1.$
$\displaystyle \\$