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This is an objective Type Questions as classified in NCERT Exemplar

Sol:

^{$\begin{array}{l}Giventhate=\frac{3}{2}\\ andfoci=\left(\pm ae,0\right)=\left(\pm 2,0\right)\\ \therefore ae=2\Rightarrow a\times \frac{3}{2}=2\Rightarrow a=\frac{4}{3}\\ Weknowthat{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow b^2=\frac{16}{9}\left(\frac{9}{4}-1\right)\Rightarrow b^2=\frac{16}{9}\times \frac{5}{4}\\ \Rightarrow b^2=\frac{20}{9}\\ So,theequationofthehyperbolais\\ \frac{x^2}{{\left(4/3\right)}^2}-\frac{y^2}{20/9}=1\Rightarrow \frac{9x^2}{16}-\frac{9y^2}{20}=1\Rightarrow \frac{x^2}{16}-\frac{y^2}{20}=\frac{1}{9}\\ \Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$}

The number of offers provided during Sanskriti University Placements has not been released yet. Check out the other placement highlights recorded during the placement drive:

The Sanskriti University the highest package offered to students during the 2025 placement drive stood at INR 54 LPA. The following table showcases the highest package offered in Sanskriti University placements: 

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$Weknowthat\text{\hspace{0.17em}Distance }$
$\begin{array}{l}betweenthefoci=2ae\\ \therefore 2ae=16\Rightarrow ae=8\\ Giventhate=\sqrt[]{2}\\ \therefore \sqrt[]{2}a=8\Rightarrow a=4\sqrt[]{2}\\ Now{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow b^2=32\left(2-1\right)\Rightarrow b^2=32\\ So,theequationofthehyperbolais\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Rightarrow \frac{x^2}{32}-\frac{y^2}{32}=1\Rightarrow x^2-y^2=32\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

As per the official record, overall 200+ recruiters participated during Sanskriti University placements 2025. The same is present in the data  tabulated below:

The number of students placed during Sanskriti University has not been released yet. Once out it will be updated on this page. Meanwhile, students can refer to the placement rate present in the table below:

Sri Muthukumaran Institute of Technology offers one undergraduate course, including BTech/B.E. The duration of the courses is four years. The college also offers MTech, MBA and MCA at the postgraduate level with a duration of two years. Students can check the table below to know detailed course-wise selection criteria:

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lengthofthelatusrectumofthehyperbola\\ =\frac{2b^2}{a}=8\Rightarrow b^2=4a\dots \left(i\right)\\ \text{\hspace{0.17em}Distance}betweenthefoci=2ae\\ Transverseaxis=2a\\ andConjugateaxis=2b\\ \therefore \frac{1}{2}\left(2ae\right)=2b\Rightarrow ae=2b\Rightarrow b=\frac{ae}{2}\dots \left(ii\right)\\ \Rightarrow b^2=\frac{a^2{e}^2}{4}\Rightarrow 4a=\frac{a^2{e}^2}{4}\left[Fromeqn.\left(i\right)\right]\\ \Rightarrow 16=a{e}^2\therefore a=\frac{16}{e^2}\\ Now{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow 4a=a^2\left(e^2-1\right)\\ \Rightarrow \frac{4}{a}=e^2-1\Rightarrow \frac{4}{16/e^2}=e^2-1\\ \Rightarrow \frac{e^2}{4}=e^2-1\Rightarrow e^2-\frac{e^2}{4}=1\Rightarrow \frac{3e^2}{4}=1\\ \Rightarrow e^2=\frac{4}{3}\therefore e=\frac{2}{\sqrt[]{3}}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Letthecoordinatesoftheverticesof\Delta ABCbe\\ A\left(x_1,y_1,z_1\right),B\left(x_2,y_2,z_2\right)andC\left(x_3,y_3,z_3\right)\\ Mid\text{\hspace{0.17em}point}ofBC=\left(1,2,-3\right)\\ \therefore 1=\frac{x_2+x_3}{2}\Rightarrow x_2+x_3=2\dots \left(i\right)\\ 2=\frac{y_2+y_3}{2}\Rightarrow y_2+y_3=4\dots \left(ii\right)\\ and-3=\frac{z_2+z_3}{2}\Rightarrow z_2+z_3=-6\dots \left(iii\right)\\ Mid\text{\hspace{0.17em}point}ofAB=\left(3,0,1\right)\\ \therefore 3=\frac{x_1+x_2}{2}\Rightarrow x_1+x_2=6\dots \left(iv\right)\\ 0=\frac{y_1+y_2}{2}\Rightarrow y_1+y_2=0\dots \left(v\right)\\ and1=\frac{z_1+z_2}{2}\Rightarrow z_1+z_2=2\dots \left(vi\right)\\ Similarly,Mid\text{\hspace{0.17em}point}ofAC=\left(-1,1,-4\right)\\ \therefore -1=\frac{x_1+x_3}{2}\Rightarrow x_1+x_3=-2\dots \left(vii\right)\\ 1=\frac{y_1+y_3}{2}\Rightarrow y_1+y_3=2\dots \left(viii\right)\\ and-4=\frac{z_1+z_3}{2}\Rightarrow z_1+z_3=-8\dots \left(ix\right)\\ Addingeq.\left(i\right),\left(iv\right)and\left(vii\right)weget,\\ 2x_1+2x_2+2x_3=2+6-2=6\\ \Rightarrow x_1+x_2+x_3=3\\ \Rightarrow 6+x_3=3\Rightarrow x_3=-3\left[?Fromeq.\left(iv\right)\right]\\ \Rightarrow x_1+2=3\Rightarrow x_1=1\left[?Fromeq.\left(i\right)\right]\\ \Rightarrow x_2-2=3\Rightarrow x_2=5\left[?Fromeq.\left(vii\right)\right]\\ So,x_1=1,x_2=5and{x}_3=-3\\ Similarly,Addingeq.\left(ii\right),\left(v\right)and\left(viii\right)weget,\\ 2\left(y_1+y_2+y_3\right)=4+0+2=6\\ \Rightarrow y_1+y_2+y_3=3\end{array}$