If the locus of the point, whose distances from the point (2,1) and (1,3) are in the ratio…

Question

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 :

Accepted Answer

$$\begin{aligned}
& \therefore \frac{(h-2)^2+(k-1)^2}{(h-1)^2+(k-3)^2}=\frac{25}{16} \
& \frac{h^2+k^2-4 h-2 k+5}{h^2+k^2-2 h-6 k+10}=\frac{25}{16} \
\end{aligned}$$

Replacing $$h \rightarrow x$$ and $$k \rightarrow y$$.

$$\begin{aligned}
& 16 x^2+16 y^2-64 x-32 y+80 \
& =25 x^2+25 y^2-50 x-150 y+250 \
& 9 x^2+9 y^2+14 x-118 y+170=0 \
& \Rightarrow a=9, b=9, c=0, d=14, e=-118 \
& a^2+2 b+3 c+4 d+e \
& 81+18+56-118 \
& \Rightarrow 37
\end{aligned}$$

Solutions · 1