If x^2-y^2+2hxy+2gx+2fy+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x+2y+7=0 and 2x-y+8=0, then the…
JEE Main 2024 — Mathematics Coordinate Geometry
2024mcqmedium
If x2−y2+2hxy+2gx+2fy+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x+2y+7=0 and 2x−y+8=0, then the value of g+c+h−f equals
Official previous-year question
Held on 30 Jan 2024 · Verified 6 Jul 2026.
Options
A
14
B
6
C
8
D
29
Did you get this right?
Sign in to track your attempts and accuracy.
Solution
Let P(x,y) be the point whose distance from x+2y+7=0 and 2x−y+8=0 is equal.
⇒5x+2y+7=±52x−y+8
⇒x2+4y2+49+4xy+28y+14x=4x2+y2+64−4xy−16y+32x
⇒−3x2+3y2−15+8xy+44y−18x=0
⇒3x2−3y2−8xy+18x−44y+15=0
⇒x2−y2−38xy+6x−344y+5=0
Now, on comparing x2−y2+2hxy+2gx+2fy+c=0 with above equation we get,
⇒h=3−4,g=3,f=−322,c=5
⇒g+c+h−f=3+5−34+322
⇒g+c+h−f=14
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.