Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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Sol:

$\begin{array}{l}Equationofhyperbolais\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \text{\hspace{0.17em}Distance}betweenthefoci=2ae\\ 2ae=16\Rightarrow ae=8\\ \Rightarrow a*\sqrt[]{2}=8\Rightarrow a=\frac{8}{\sqrt[]{2}}=4\sqrt[]{2}\left[?e=\sqrt[]{2}\right]\\ Now,b^2=a^2\left(e^2-1\right)\left[Forhyperbola\right]\\ \Rightarrow b^2={\left(4\sqrt[]{2}\right)}^2\left(2-1\right)\\ \Rightarrow b^2=32\\ a=4\sqrt[]{2}\Rightarrow a^2=32\\ Hence,therequiredequationis\frac{x^2}{32}-\frac{y^2}{32}=1\Rightarrow x^2-y^2=32.\end{array}$

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Sol:

$\begin{array}{l}Giventhat{y}^2=4x\dots \left(i\right)\\ andy=mx+1\dots \left(ii\right)\\ Fromeqn.\left(i\right)and\left(ii\right)weget\\ {\left(mx+1\right)}^2=4x\\ \Rightarrow m^2{x}^2+1+2mx-4x=0\\ \Rightarrow m^2{x}^2+\left(2m-4\right)x+1=0\\ Applyingconditionof\text{\hspace{0.17em}tangency}wehave\\ {\left(2m-4\right)}^2-4m^2*1=0\\ \Rightarrow 4m^2+16-16m-4m^2=0\\ \Rightarrow -16m=-16\Rightarrow m=1\\ Hence,therequiredvalueofmis1.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:\\ Vertex=\left(0,4\right)andFocus=\left(0,2\right)\\ LetP\left(x,y\right)beany\text{\hspace{0.17em}point}ontheparabola.PBisperpendiculartothedirectrix.\\ Accordingtothedefinitionofparabola,wehave\\ PF=PB\\ \Rightarrow \sqrt[]{{\left(x-0\right)}^2+{\left(y-2\right)}^2}=\left|\frac{0+y-6}{\sqrt[]{0+1}}\right|\left[?Equationofdirectrixisy=6\right]\\ \Rightarrow \sqrt[]{x^2+{\left(y-2\right)}^2}=\left(y-6\right)\\ Squaringbothsides,wehave\\ x^2+y^2+4-4y=y^2+36-12y\\ \Rightarrow x^2-4y+12y-32=0\\ \Rightarrow x^2+8y-32=0\\ Hence,therequiredequationis{x}^2+8y=32.\end{array}$

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Sol:

$\begin{array}{l}Givenparabolais{y}^2=4ax\\ LetP\left(a{t}^2,2at\right)beanyontheparabola.\\ In\Delta POA,\\ tan\theta =\frac{2at}{a{t}^2}=\frac{2}{t}\\ \Rightarrow t=\frac{2}{tan\theta }\\ \Rightarrow t=2cot\theta \dots \left(i\right)\\ OP=\sqrt[]{{\left(a{t}^2-0\right)}^2+{\left(2at-0\right)}^2}\\ =\sqrt[]{a^2{t}^4+4a^2{t}^2}\\ =at\sqrt[]{t^2+4}\\ =a*2cot\theta \sqrt[]{4{}^{}+4}\left[?t=2cot\theta \right]\\ =2acot\theta .2\sqrt[]{co{t}^2\theta +1}=4acot\theta cosec\theta \\ =4a.\frac{cos\theta }{sin\theta }.\frac{1}{sin\theta }=\frac{4acos\theta }{si{n}^2\theta }\\ Hence,therequiredlength=\frac{4acos\theta }{si{n}^2\theta }.\end{array}$

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Sol:

$\begin{array}{l}Givenparabolais{y}^2=8x\dots \left(i\right)\\ Comparingwiththeequationofparabola{y}^2=4ax\\ 4a=8\Rightarrow a=2\\ Nowfoca\mathrm{lDistance}=\left|x+a\right|\\ \Rightarrow \left|x+a\right|=4\\ \Rightarrow \left(x+a\right)=\pm 4\Rightarrow x+2=\pm 4\\ \Rightarrow x=4-2=2andx=-6\\ Butx\ne -6\therefore x=2\\ Putx=2inequation\left(i\right)weget\\ y^2=8*2=16\\ \therefore y=\pm 4\\ So,thecoordinatesofthe\text{points}are\left(2,4\right),\left(2,-4\right).\\ Hence,therequiredcoordinatesare\left(2,4\right),\left(2,-4\right).\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationofanellipseis\frac{x^2}{36}+\frac{y^2}{20}=1\\ Here,a^2=36\Rightarrow a=6and{b}^2=20\Rightarrow b=2\sqrt[]{5}\\ Weknowthat{b}^2=a^2\left(1-e^2\right)\\ \Rightarrow 20=36\left(1-e^2\right)\\ \Rightarrow 1-e^2=\frac{20}{36}\Rightarrow e^2=1-\frac{20}{36}=\frac{16}{36}\\ \Rightarrow e=\frac{4}{6}=\frac{2}{3}\\ Now\text{\hspace{0.17em}Distance}betweenthedirectricesis\\ \frac{a}{e}-\left(-\frac{a}{e}\right)=\frac{a}{e}+\frac{a}{e}=\frac{2a}{e}\\ =2*\frac{6}{2/3}=2*6*\frac{3}{2}=18\\ Hence,therequired\text{\hspace{0.17em}distance}=18.\end{array}$

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Sol:

$\begin{array}{l}Equationofanellipseis\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\dots \left(i\right)\\ Giventhat,e=\frac{2}{3}\\ andlatusrectum=\frac{2b^2}{a}=5\\ \Rightarrow b^2=\frac{5}{2}a\dots \left(ii\right)\\ Weknowthat{b}^2=a^2\left(1-e^2\right)\\ \Rightarrow \frac{5}{2}a=a^2\left(1-\frac{4}{9}\right)\\ \Rightarrow \frac{5}{2}=a*\frac{5}{9}\Rightarrow a=\frac{9}{2}\Rightarrow a^2=\frac{81}{4}\\ and{b}^2=\frac{5}{2}*\frac{9}{2}=\frac{45}{4}\\ Hence,therequiredequationofellipseis\\ \frac{x^2}{81/4}+\frac{y^2}{45/4}=1\Rightarrow \frac{4}{81}{x}^2+\frac{4}{45}{y}^2=1\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Equationofanellipseis\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \text{Distance\hspace{0.17em}}betweenitsfoci=ae+ae=2ae\\ \therefore 2ae=10\\ \Rightarrow ae=5\Rightarrow a*\frac{5}{8}=5\Rightarrow a=8\\ Now{b}^2=a^2\left(1-e^2\right),whereeistheeccentricity\\ \Rightarrow b^2=64\left(1-\frac{25}{64}\right)\\ \Rightarrow b^2=64*\frac{39}{64}\Rightarrow b^2=39\\ So,thelengthofthelatusrectum=\frac{2b^2}{a}=\frac{2*39}{8}=\frac{39}{4}\\ Hence,lengthofthelatusrectum=\frac{39}{4}.\end{array}$