Maths

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}yandzcoordinates.\\ Hence,thevalueofthefillerisyandzcoordinates.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}z-coordinates.\\ Hence,thevalueofthefillerisz-coordinates.\end{array}$

$\int {\left(\frac{x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)}^{2}dx=\int \left(\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)\cdot \frac{x\mathrm{c}\mathrm{o}\mathrm{s}?xdx}{(x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x{)}^2}$

$\begin{array}{rr}& =\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\left(-\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)+\int \left(\frac{\mathrm{c}\mathrm{o}\mathrm{s}?x+x\mathrm{s}\mathrm{i}\mathrm{n}?x}{{\mathrm{c}\mathrm{o}\mathrm{s}}^2?x}\right)\left(\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)dx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\int {\mathrm{s}\mathrm{e}\mathrm{c}}^2?xdx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\mathrm{t}\mathrm{a}\mathrm{n}?x+C\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Theequationofz-axisare,x=0andy=0.\\ Hence,thevalueofthefillerisx=0andy=0.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Onthez-axis,x=0andy=0.\\ \therefore Therequiredcoordinateisintheformof\left(0,0,z\right).\\ Hence,thevalueofthefilleris\left(0,0,z\right).\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Theequationofyz-planeisx=0.\\ Hence,thevalueofthefillerisx=0.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Weknowthatyzplane,x=0\\ So,thecoordinatesoftherequired\text{\hspace{0.17em}point}is\left(0,y,z\right).\\ Hence,thevalueofthefilleris\left(0,y,z\right).\end{array}$

Since $\left(\mathrm{3,3}\right)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{9}{a^2}-\frac{9}{b^2}=1$

Now, normal at $\left(\mathrm{3,3}\right)$ is $y-3=-\frac{a^2}{b^2}(x-3)$ ,

which passes through $\left(\mathrm{9,0}\right)\Rightarrow b^2=2a^2$

So,  ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=3$

Also,  ${\mathrm{a}}^2=\frac{9}{2}$

(From (i) and (ii)

Thus,  $\left({\mathrm{a}}^2,{\mathrm{e}}^2\right)=\left(\frac{9}{2},3\right)$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Eight.\\ Hence,thevalueofthefilleriseight.\end{array}$

$\left(\frac{\mathrm{K}-2}{\mathrm{h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1\Rightarrow \mathrm{K}=2\mathrm{h}$

$?[?\mathrm{A}\mathrm{B}\mathrm{C}]=5\sqrt[]{5}$

$\Rightarrow \frac{1}{2}\left(\sqrt[]{5}\right)\sqrt[]{(\mathrm{h}-1{)}^2+(\mathrm{K}-2{)}^2}=5\sqrt[]{5}$