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Let AB be a line segment of length 4 with A on the line $\mathrm{y = 2x \; and\; B}$ on the line $\mathrm{y = x}$ . Find k such that the locus of the mid-points of all such line segments is the curve $\mathrm{25x^{2} + 13y^{2} - 36xy - k = 0.}$
Option: 1
$\mathrm{K=-4}$

Option: 2
$\mathrm{K=3}$

Option: 3
$\mathrm{K=2}$

Option: 4
$\mathrm{K=4}$

Answers (1)

best_answer

$\text{Let}\: \mathrm{B} \equiv(\mathrm{h}, \mathrm{h}) \: \text{and} \: \mathrm{A} \equiv(\mathrm{k}, 2 \mathrm{k})$

$\text{Let P} \: (\alpha, \beta)$ be the mid point of AB

So, $\mathrm{2 \alpha=h+k \: \text { and }\: 2 \beta=h+2 k}$

$\mathrm{\Rightarrow \mathrm{k}=2 \beta-2 \alpha \text { and } \mathrm{h}=4 \alpha-2 \beta}$

Given $\mathrm{A B=4 \text { or } A B^2=16}$

$\mathrm{\begin{aligned} & \Rightarrow \mathrm{(h-k)^2+(h-2 k)^2=16} \\ \\& \Rightarrow(6 \alpha-4 \beta)^2+(8 \alpha-6 \beta)^2=16 \end{aligned}}$

$\begin{aligned} & \Rightarrow 100 \alpha^2+52 \beta^2-144 \alpha \beta=16 \\ \\& \Rightarrow 25 \alpha^2+13 \beta^2-36 \alpha \beta=4 \end{aligned}$

$\therefore$ locus of $\mathrm{P(\alpha, \beta) \text { is } 25 x^2+13 y^2-36 x y-4=0}$

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Shailly goel

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