JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 8)
If $$x^2-y^2+2 h x y+2 g x+2 f y+c=0$$ is the locus of a point, which moves such that it is always equidistant from the lines $$x+2 y+7=0$$ and $$2 x-y+8=0$$, then the value of $$g+c+h-f$$ equals
8
14
29
6
Explanation
Cocus of point $$\mathrm{P}(\mathrm{x}, \mathrm{y})$$ whose distance from Gives $$X+2 y+7=0$$ & $$2 x-y+8=0$$ are equal is $$\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}$$
$$(x+2 y+7)^2-(2 x-y+8)^2=0$$
Combined equation of lines
$$\begin{aligned} & (x-3 y+1)(3 x+y+15)=0 \\ & 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\ & x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \\ & x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\ & h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \\ & g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14 \end{aligned}$$
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