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QJEEM24Ma5151: JEE Main 2024 Mathematics - Class 11 JEEM24Ma5151 Solution: If is the locus of a point, which moves such that it is always…

If x2−y2+2hxy+2gx+2fy+c=0x^2-y^2+2 h x y+2 g x+2 f y+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x+2y+7=0x+2 y+7=0 and 2x−y+8=02 x-y+8=0 , then the value of g+c+h−fg+c+h-f equals

Multiple Choice Options

  • A 8
  • B 14
  • C 29
  • D 6

Step-by-Step Verified Solution

Cocus of point P(x,y)\mathrm{P}(\mathrm{x}, \mathrm{y}) whose distance from Gives X+2y+7=0X+2 y+7=0 & 2x−y+8=02 x-y+8=0 are equal is x+2y+75=±2x−y+85\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}

(x+2y+7)2−(2x−y+8)2=0(x+2 y+7)^2-(2 x-y+8)^2=0

Combined equation of lines

amp;(x−3y+1)(3x+y+15)=0amp;3x2−3y2−8xy+18x−44y+15=0amp;x2−y2−83xy+6x−443y+5=0amp;x2−y2+2hxy+2gx2+2fy+c=0amp;h=43,g=3,f=−223,c=5amp;g+c+h−f=3+5−43+223=8+6=14\begin{aligned}& (x-3 y+1)(3 x+y+15)=0 \\& 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\& x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \\& x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\& h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \\& g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14\end{aligned}

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