Mathematics · Straight lines

JEE Main 2025 — 23 January, Evening Shift — Question 5

A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$ ,

respectively. If the locus of the point $P$ , that divides the rod AB internally in the ratio $2: 1$ is

$9\left(x^{2}+\alpha y^{2}+\beta x y+\gamma x+28y\right)-76=0$ , then $\alpha-\beta-\gamma$ is equal to :

Answer: B

$\mathrm{h}=\frac{2\beta+\alpha}{3}$

$\mathrm{k}=\frac{-4+\alpha+2}{3}$

$\alpha=3 \mathrm{k}+2$

$2 \beta=3 h-a=3 h-3 k-2$

so $\mathrm{AB}=8$

$(\alpha-\beta)^{2}+(\alpha+4)^{2}=64$

$\left(3 \mathrm{k}+2-\left(\frac{3 \mathrm{~h}-3 \mathrm{k}-2}{2}\right)\right)^{2}+(3 \mathrm{k}+2+4)^{2}=64$

$\frac{(9 \mathrm{k}-3 \mathrm{~h}+6)^{2}}{4}+(3 \mathrm{k}+6)^{2}=64$

$9\left[(3 k-h+2)^{2}+4(k+2)^{2}\right]=64 \times 4$

$9\left(x^{2}+13 y^{2}-6 x y-4 x+28 y\right)=76$

$\alpha-\beta-\gamma=13+6+4=23$

Answer key and solution verified before publishing.

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