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QJEEM25Ma5718: JEE Main 2025 Mathematics - Class 11 JEEM25Ma5718 Solution: A rod of length eight units moves such that its ends and always …

A rod of length eight units moves such that its ends AA and BB always lie on the lines x−y+2=0x-y+2=0 and y+2=0y+2=0 , respectively. If the locus of the point PP , that divides the rod ABA B internally in the ratio 2:12: 1 is 9(x2+αy2+βxy+γx+28y)−76=09\left(x^2+\alpha y^2+\beta x y+\gamma x+28 y\right)-76=0 , then α−β−γ\alpha-\beta-\gamma is equal to :

Multiple Choice Options

  • A 24
  • B 22
  • C 21
  • D 23

Step-by-Step Verified Solution

JEE Main 2025 (Online) 23rd January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 3 English Explanation

amp;h=3β+α3amp;k=−4+α+23amp;α=3k+2amp;2β=3 h−a=3 h−3k−2amp; so AB=8amp;(α−β)2+(α+4)2=64amp;(3k+2−(3 h−3k−22))2+(3k+2+4)2=64amp;(9k−3 h+6)24+(3k+6)2=64amp;9[(3k−h+2)2+4(k+2)2]=64×4amp;9(x2+13y2−6xy−4x+28y)=76amp;α−β−γ=13+6+4=23\begin{aligned}& \mathrm{h}=\frac{3 \beta+\alpha}{3} \\& \mathrm{k}=\frac{-4+\alpha+2}{3} \\& \alpha=3 \mathrm{k}+2 \\& 2 \beta=3 \mathrm{~h}-\mathrm{a}=3 \mathrm{~h}-3 \mathrm{k}-2 \\& \text { 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 \mathrm{k}-\mathrm{h}+2)^2+4(\mathrm{k}+2)^2\right]=64 \times 4 \\& 9\left(\mathrm{x}^2+13 \mathrm{y}^2-6 \mathrm{xy}-4 \mathrm{x}+28 \mathrm{y}\right)=76 \\& \alpha-\beta-\gamma=13+6+4=23\end{aligned}

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