JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 20)
If the locus of the point, whose distances from the point $$(2,1)$$ and $$(1,3)$$ are in the ratio $$5: 4$$, is $$a x^2+b y^2+c x y+d x+e y+170=0$$, then the value of $$a^2+2 b+3 c+4 d+e$$ is equal to :
37
$$-27$$
437
5
Explanation
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$$\begin{aligned} & \therefore \frac{(h-2)^2+(k-1)^2}{(h-1)^2+(k-3)^2}=\frac{25}{16} \\ & \frac{h^2+k^2-4 h-2 k+5}{h^2+k^2-2 h-6 k+10}=\frac{25}{16} \\ \end{aligned}$$
Replacing $$h \rightarrow x$$ and $$k \rightarrow y$$.
$$\begin{aligned} & 16 x^2+16 y^2-64 x-32 y+80 \\ & =25 x^2+25 y^2-50 x-150 y+250 \\ & 9 x^2+9 y^2+14 x-118 y+170=0 \\ & \Rightarrow a=9, b=9, c=0, d=14, e=-118 \\ & a^2+2 b+3 c+4 d+e \\ & 81+18+56-118 \\ & \Rightarrow 37 \end{aligned}$$
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