Question 67

If the locus of the point, whose distances from the point $$(2, 1)$$ and $$(1, 3)$$ are in the ratio $$5 : 4$$, is $$ax^2 + by^2 + cxy + dx + ey + 170 = 0$$, then the value of $$a^2 + 2b + 3c + 4d + e$$ is equal to :

$$\frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2} = 25/16$$.

$$16[(x-2)^2+(y-1)^2] = 25[(x-1)^2+(y-3)^2]$$.

$$16(x²-4x+4+y²-2y+1) = 25(x²-2x+1+y²-6y+9)$$.

$$16x²-64x+80+16y²-32y = 25x²-50x+250+25y²-150y$$.

$$9x²+9y²+14x-118y+170 = 0$$. So a=9, b=9, c=0, d=14, e=-118.

a²+2b+3c+4d+e = 81+18+0+56-118 = 37.

The correct answer is Option (1): 37.

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