Mathematics · Straight lines

JEE Main 2024 — 6 April, Shift 2 — Question 5

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:

Answer: C

let $\mathrm{P}(\mathrm{x}, \mathrm{y})$

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

$9 x^{2}+9 y^{2}+14 x-118 y+170=0$

$\mathrm{a}^{2}+2 \mathrm{~b}+3 \mathrm{c}+4 \mathrm{~d}+\mathrm{e}$

$=81+18+0+56-118$

$=155-118$

$=37$

Answer key and solution verified before publishing.

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