Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The foot of the perpendicular from a point on the circle <img alt="\mathrm{ x^{2}+y^{2}=1, \mathrm{z}=0}" src="/latex-image/?%5Cmathrm%7B%20x%5E%7B2%7D&plus;y%5E%7B2%7D%3D1%2C%20%5Cmathrm%

The foot of the perpendicular from a point on the circle \mathrm{ x^{2}+y^{2}=1, \mathrm{z}=0} to the plane \mathrm{ 2 x+3 y+z=6} lies on which one of the following curves ?

Option: 1

\mathrm{(6 x+5 y-12)^{2}+4(3 x+7 y-8)^{2}=1, z=6-2 x-3 y}


Option: 2

\mathrm{(5 x+6 y-12)^{2}+4(3 x+5 y-9)^{2}=1, z=6-2 x-3 y}


Option: 3

\mathrm{(6 x+5 y-14)^{2}+9(3 x+5 y-7)^{2}=1, \quad z=6-2 x-3 y}


Option: 4

\mathrm{(5 x+6 y-14)^{2}+9(3 x+7 y-8)^{2}=1, z=6-2 x-3 y}


Answers (1)

\begin{aligned} & \mathrm{\frac{h-\cos \theta}{2}=\frac{k-\sin \theta}{3}=\frac{\omega-0}{1}=\frac{-1(2 \cos \theta+3 \sin \theta-6)}{14}}\\ & \mathrm{\frac{h-\cos \theta}{2}=\frac{-(2 \cos \theta+3 \sin \theta-6)}{14}}\\ & \mathrm{h=\frac{\cos \theta-2(2 \cos \theta+3 \sin \theta-6)}{14}}\\ & \mathrm{n=\frac{10 \cos \theta-6 \sin \theta+12}{14}}\\ & \mathrm{1<=\sin \theta-\frac{3}{14}(2 \cos \theta+3 \sin \theta-6)}\\ &\mathrm{k=\frac{5 \sin \theta-6 \cos \theta+18}{14}}\\ & \text{ Elementary } \operatorname{Sin} \theta \text{ and } \operatorname{Cos} \theta\\ &\mathrm{(5 h+6 k-12)^{2}+4(3 h+5 k-9)^{2}=1}\\ \end{aligned}

 

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question