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

Sol:

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

Yes, IFIM College of Law admissions 2025-26 admissions for BBA LLB, and other courses are currently open. Students can visit the official website to learn more and apply online. IFIM Law has been ranked 1st in the Category of Top Eminent Law School by the GHRDC Survey 2024 and ranked 6th among the Top Private Law Schools in Karnataka by IIRF 2024.

Yes, candidates can joint the Indian School of Business Management and Administration, Chennai after Class 12. Candidates are required to fill out the application form for the courses they are eligible to. Candidates who have cleared Class 12 can apply to the Certificate, Diploma, and UG courses. 

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

Sol:

$\begin{array}{l}Givenequationofanellipseis\\ 9x^2+25y^2=225\\ \Rightarrow \frac{9}{225}{x}^2+\frac{25}{225}{y}^2=1\\ \Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1\\ Here,a=5andb=3\\ Now{b}^2=a^2\left(1-e^2\right),whereeistheeccentricity\\ \Rightarrow 9=25\left(1-e^2\right)\\ \Rightarrow 1-e^2=\frac{9}{25}\Rightarrow e^2=1-\frac{9}{25}=\frac{16}{25}\\ \therefore e=\frac{4}{5}\\ Nowfoci=\left(\pm ae,0\right)=\left(\pm 5\times \frac{4}{5},0\right)=\left(\pm 4,0\right)\\ Hence,eccentricityis\frac{4}{5},foci=\left(\pm 4,0\right).\end{array}$

NIELIT Ajmer offers BTech and BTech Integrated courses based on JEE Main scores. Candidates having valid JEE Main ranks will be allowed to take part in JoSAA counselling rounds. Considering the NIELIT Ajmer JEE Main cutoff 2025 for the Round 4 seat allotment results, the B.Tech. in Computer Science and Engineering (Internet of Things, Cyber Security including Block Chain Technology closing rank for the General AI category stood at 56625. It is the only course for which the institute releases the cutoffs. Check out the NIELIT Ajmer CSE year-wise Round 4 closing ranks from the table below:

Candidates seeking admission to the Global BIFS Academy, must fulfil the college's eligibility criteria. For certificate course, applicants must have completed Class 12 and graduation with a 50% aggregate. The selection process for all courses is based on students merit, i.e., Career Counsellor Screening.

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

Sol:

Sol:

$\begin{array}{l}Lettheequationofanellipseis\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ Lengthofmajoraxis=2a\\ Lengthof\text{\hspace{0.17em}minor}axis=2b\\ andthelengthoflatusrectum=\frac{2b^2}{a}\\ Accordingtothequestion,wehave\\ \frac{2b^2}{a}=\frac{2b}{2}\Rightarrow b=\frac{a}{2}\\ Now{b}^2=a^2\left(1-e^2\right),whereeistheeccentricity\\ \Rightarrow b^2=4b^2\left(1-e^2\right)\\ \Rightarrow 1=4\left(1-e^2\right)\Rightarrow 1-e^2=\frac{1}{4}\Rightarrow e^2=\frac{3}{4}\\ \therefore e=\pm \frac{\sqrt[]{3}}{2}\\ So,e=\frac{\sqrt[]{3}}{2}\left[?eisnot\left(-\right)\right]\\ Hence,therequiredvalueofeccentricityis\frac{\sqrt[]{3}}{2}.\end{array}$

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

Sol:

$\begin{array}{l}Givenequationofthecircleis\\ x^2+y^2-6x+12y+15=0\dots \left(i\right)\\ Centre=\left(-g,-f\right)=\left(3,-6\right)\left[\begin{array}{l}?2g=-6\Rightarrow g=-3\\ 2f=12\Rightarrow f=6\end{array}\right]\\ \text{\hspace{0.17em}Since}thecircleisconcentricwiththegivencircle\\ \therefore Centre=\left(3,-6\right)\\ Nowlettheradiusofthecircleisr\\ \therefore r=\sqrt[]{g^2+f^2-c}=\sqrt[]{9+36-15}=\sqrt[]{30}\\ Areaofthegivencircle\left(i\right)=\pi r^2=30\pi sq.unit\\ Areaoftherequiredcircle=2\times 30\pi =60\pi sq.unit\\ If{r}_{1}betheradiusoftherequiredcircle\\ \pi r_1^2=60\pi \Rightarrow r_1^2=60\\ So,therequiredequationofthecircleis\\ {\left(x-3\right)}^2+{\left(y+6\right)}^2=60\\ \Rightarrow x^2+9-6x+y^2+36+12y-60=0\\ \Rightarrow x^2+y^2-6x+12y-15=0\\ Hence,therequiredequationis{x}^2+y^2-6x+12y-15=0.\end{array}$

Amity University Kolkata admissions have commenced. Candidates must visit the official website to apply for the same. Candidates must fill out the online application form to register for candidature. After a successful application submission, candidates will be shortlisted based on the eligibility criteria provided. For final admission, shortlisted candidates must pay the fees and confirm the seat. 

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

Sol:

$\begin{array}{l}Givencircleis{x}^2+y^2=16\\ Perpendicularfromtheorigintothegivenliney=\sqrt[]{3}x+kisequaltotheradius.\\ \therefore 4=\left|\frac{0-0-k}{\sqrt[]{{\left(1\right)}^2+{\left(\sqrt[]{3}\right)}^2}}\right|=\left|\frac{-k}{4}\right|\\ \Rightarrow 4=\pm \frac{k}{2}\Rightarrow k=\pm 8\\ Hence,therequiredvaluesofkare\pm 8.\end{array}$

Tezpur University has released its B.Arch., BTech and BTech Integrated courses for the All India quotas. For All India, BTech cutoffs were released by the university. Considering the Tezpur University JEE Main cutoff 2025 for the Round 4 seat allotment results, yes, candidates scoring a rank of 50000 can get admission into the university. Candidates can even fetch BTech in CSE programme with this much of rank. The BTech in Computer Science Engineering cutoff stood at,  60587 for the General All India quota candidates.

Candidates have to meet the merit requirements to get admission in UG courses offered at Acharya Institute of Health Sciences. Those who are selected have to report at the campus for document verification and also pay the course fee to confirm their seat. Below are the documents required for UG admission:

• Class 10 and Class 12 mark card
• Transfer Certificate
• Caste Certificate
• Fitness Certificate / Medical Certificate
• Age Proof Certificate (in case DOB is not mentioned in the testimonials)
• Other related documents, if any (Sports. Co-Curricular etc.)
• Photographs – 10 stamp size / 10 passport size
• 2 sets of copies of all the testimonials

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

Sol:

$\begin{array}{l}Givenequationsare\\ 3x+y=14\dots \left(i\right)\\ and2x+5y=18\dots \left(ii\right)\\ Fromeq.\left(i\right)weget\\ y=14-3x\dots \left(iii\right)\\ Puttingthevalueofyineq.\left(ii\right)weget\\ \Rightarrow 2x+5\left(14-3x\right)=18\\ \Rightarrow 2x+70-15x=18\\ \Rightarrow -13x=-70+18\\ \Rightarrow -13x=-52\Rightarrow x=4\\ Fromeq.\left(iii\right)weget,y=14-3\times 4=2\\ \therefore \text{\hspace{0.17em}point}o\mathrm{fintersection}is\left(4,2\right)\\ Now,radiusr=\sqrt[]{{\left(4-1\right)}^2+{\left(2+2\right)}^2}=\sqrt[]{{\left(3\right)}^2+{\left(4\right)}^2}=\sqrt[]{9+16}=5\\ So,theequationofcircleis\\ {\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \Rightarrow {\left(x-1\right)}^2+{\left(y+2\right)}^2={\left(5\right)}^2\\ \Rightarrow x^2-2x+1+y^2+4y+4=25\\ \Rightarrow x^2+y^2-2x+4y-20=0\\ Hence,therequiredequationis{x}^2+y^2-2x+4y-20=0.\end{array}$