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%5

If the equation  \mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0}  represents a pair of straight lines equidistant from the origin, then  \mathrm{f}^4-\mathrm{g}^4=\mathrm{k}\left(\mathrm{bf}^2-\mathrm{af}^2\right)  where \mathrm{k}=

Option: 1

\frac{c}{2}


Option: 2

2c


Option: 3

c


Option: 4

None of these


Answers (1)

best_answer

Given equation is  \mathrm{ax}^2+2 \mathrm{hxy}+\mathrm{by}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0
Let the lines represented by (i) be  l \mathrm{x}+\mathrm{my}+\mathrm{n}=0  and  \mathrm{l^{\prime} \mathrm{x}+\mathrm{m}^{\prime} \mathrm{y}+\mathrm{n}^{\prime}=0}
So that we have  \mathrm{l l^{\prime}=\mathrm{a}, \mathrm{mm}^{\prime}=\mathrm{b}, \mathrm{nn}^{\prime}=\mathrm{c} . l \mathrm{~m}^{\prime}+l^{\prime} \mathrm{m}=2 \mathrm{~h}}\mathrm{l n^{\prime}+l^{\prime} n=2 g, m^{\prime}+m^{\prime} n=2 f}..........(ii)

since they are equidistant from the origin,\quad \therefore \frac{|\mathrm{n}|}{\sqrt{1^2+\mathrm{m}^2}}=\frac{\left|\mathrm{n}^{\prime}\right|}{\sqrt{\mathrm{l}^{\prime 2}+\mathrm{m}^{\prime 2}}} 

or \mathrm{n}^2 l^{12}+\mathrm{n}^2 \mathrm{~m}^{\prime 2}=\mathrm{n}^{1 / 2}+\mathrm{n}^{12} \mathrm{~m}^2

\left(\mathrm{n} l^{\prime}-\mathrm{n}^{\prime} l\right)\left(\mathrm{n} l^{\prime}+\mathrm{n}^{\prime} l\right)=\left(\mathrm{mn}^{\prime}-\mathrm{m}^{\prime} \mathrm{n}\right)\left(\mathrm{mn}^{\prime}+\mathrm{m}^{\prime} \mathrm{n}\right)

or \left(\mathrm{n} l^{\prime}-\mathrm{n}^{\prime} l\right)^2\left(\mathrm{n} l^{\prime}+\mathrm{n}^{\prime} l\right)^2=\left(\mathrm{mn}^{\prime}-\mathrm{m}^{\prime} \mathrm{n}\right)^2\left(\mathrm{mn}^{\prime}+\mathrm{m}^{\prime} \mathrm{n}\right)^2  or 

\text { or } \left.\left[\mathrm{n} l^{\prime}+\mathrm{n}^{\prime} l\right)^2-4 l l^{\prime} \mathrm{nn}^{\prime}\right]\left(\mathrm{n} l^{\prime}+\mathrm{n}^{\prime} l\right)^2

=\left[\left(\mathrm{mn}^{\prime}+\mathrm{m}^{\prime} \mathrm{n}\right)^2-4 \mathrm{~mm}^{\prime} \mathrm{nn}^{\prime}\right]\left(\mathrm{n} l^{\prime}+\mathrm{n}^{\prime} l\right)^2    or 

\mathrm{\left(4 g^2-4 a c\right) \cdot 4 g^2=\left(4 f^2-4 b c\right) 4 f^2 \text { or } f^4-g^4=c\left(b f^2-a f^2\right) \text {. }}

Posted by

Riya

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question