Question bank/JEE Main/Mathematics

JEE MainMathematicsModerateCircles

Chord of the curve 4x2+y2−x+4y=04x^{2} + y^{2} - x + 4y = 0which subtend a
right angle at the origin pass through a fixed point whose co-ordinate
are
oØ 4x2+y2−x+4y=04x^{2} + y^{2} - x + 4y = 0 dh og thok tks ewy fcUnq ij ledks.k
vUrfjr djrh gS,d fuf'pr fcUnq

Show answer and solution
Let equation of chord lx+my=n…….(i)\mathcal{l}x + my = n\ldots\ldots.(i)
ekuk thok dk lehdj.k lx+my=n…….(i)\mathcal{l}x + my = n\ldots\ldots.(i)
lx+myn=1\frac{\mathcal{l}x + my}{n} = 1
c1:4x2+y2−x+4y=0c_{1}:4x^{2} + y^{2} - x + 4y = 0
4x2+y2−x(lx+myn)+4y(lx+myn)=04x^{2} + y^{2} - x\left( \frac{\mathcal{l}x + my}{n} \right) + 4y\left( \frac{lx + my}{n} \right) = 0
 coefficient of x2+ coefficient of y2=0\ coefficient\ of\ x^{2} + \ coefficient\ of\ y^{2} = 0
x2x^{2} dk xq.kkad + y2y^{2} dk xq.kkad = 0
(4−ln)+(1+4mn)=0\left( 4 - \frac{\mathcal{l}}{n} \right) + \left( 1 + \frac{4m}{n} \right) = 0
4n−l+n+4m=04n\mathcal{- l +}n + 4m = 0
l−4m=5n\mathcal{l -}4m = 5n
From (i) and (ii)\ (i)\ and\ (ii)
xl=y−4=15\frac{x}{\mathcal{l}} = \frac{y}{- 4} = \frac{1}{5}
x=15,y=−45x = \frac{1}{5},y = \frac{- 4}{5}
passes through (15,−45)\left( \frac{1}{5},\frac{- 4}{5} \right)

Source: Grass-G / M-25 Mathematics Paper 7 Jan.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans