Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the equation <img alt="\mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba%20x%5E2&plus;2%20h%20x%20y&plus;b%20y%5E2&plus

If the equation \mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0} represents two straight lines, the product of the perpendiculars drawn from the origin to these lines is \mathrm{\frac{c}{\sqrt{(a-b)^2+k h^2}}}, where \mathrm{k =}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

We have,  \mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0}  .....(1)

Let,  \mathrm{l \mathrm{x}+\mathrm{my}+\mathrm{n}=0}   .....(2)

and, \mathrm{l^{\prime} \mathrm{x}+\mathrm{m}^{\prime} \mathrm{y}+\mathrm{n}^{\prime}=0}   .....(3)

be the lines represented by the given equation. Then (l \mathrm{x}+\mathrm{my}+\mathrm{n})\left(l^{\prime} \mathrm{x}+\mathrm{m}^{\prime} \mathrm{y}+\mathrm{n}^{\prime}\right) \equiv \mathrm{xx}^2+2 \mathrm{hxy}+\mathrm{by}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}

comparing the coefficients of like terms, we obtain l l^{\prime}=\mathrm{a}, \mathrm{mm}^{\prime}=\mathrm{b},

l \mathrm{~m}^{\prime}+l^{\prime} \mathrm{m}=2 \mathrm{~h}, \begin{gathered} l^{\prime}+l^{\prime} \mathrm{n}=2 \mathrm{~g}, \quad \mathrm{mn}^{\prime}+\mathrm{m}^{\prime} \mathrm{n}=2 f \\ \text { and, } \mathrm{nn}^{\prime}=\mathrm{c} \end{gathered}

Let \mathrm{p}_1 and \mathrm{p}_2 be the lengths of perpendicular from the origin on (ii) and (iii) respectively. Then,

\mathrm{\mathrm{p}_1=\frac{\mathrm{n}}{\sqrt{l^2+\mathrm{m}^2}} \text { and } \mathrm{p}_2=\frac{\mathrm{n}^{\prime}}{\sqrt{l^{\prime 2}+\mathrm{m}^{\prime 2}}}}  \mathrm{\Rightarrow \mathrm{p}_1 \mathrm{p}_2=\frac{\mathrm{nn}^{\prime}}{\sqrt{\left(l^2+\mathrm{m}^2\right)\left(l^{\prime 2}+\mathrm{m}^{\prime 2}\right)}}}

\mathrm{=\frac{\mathrm{nn}^{\prime}}{\sqrt{\left(l l^{\prime}\right)^2+\left(\mathrm{mm}^{\prime}\right)^2+\left(l \mathrm{~m}^{\prime}\right)^2+\left(l^{\prime} \mathrm{m}\right)^2}}=\frac{\mathrm{nn}^{\prime}}{\sqrt{\left(l l^{\prime}-\mathrm{mm}^{\prime}\right)^2+\left(l \mathrm{~m}^{\prime}+l^{\prime} \mathrm{m}\right)^2}}}

\mathrm{=\sqrt{\frac{c}{(a-b)^2+4 h^2}}}

\mathrm{[l l^{\prime}=\mathrm{a}, \mathrm{mm}^{\prime}=\mathrm{b} \text { and } l \mathrm{~m}^{\prime}+l^{\prime} \mathrm{m}=2 \mathrm{~h}]}

 

 

Posted by

Nehul

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 Mains 2026 Exam City Slip LIVE: NTA session 2 test centre allotment soon; admit card date
anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 city slip soon; admit card, exam dates, here's how to prepare

Latest Question