Pair of Lines JEE Advanced previous year questions with solutions

2 solved JEE Advanced questions on Pair of Lines, free to read — no sign-in needed.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   Let $S$ denote the locus of the point of intersection of the pair of lines $4 x-3 y=12 \alpha$ $4 \alpha x+3 \alpha y=12$ where $\alpha$ varies over the set of non-zero real numbers. Let $T$ be the tangent to $S$ passing through the points $(p, 0)$ and $(0, q), q\gt 0$, and parallel to the line $4 x-\frac{3}{\sqrt{2}} y=0$. Then the value of $p q$ is
   1. A.$-6 \sqrt{2}$
   2. B.$-3 \sqrt{2}$
   3. C.$-9 \sqrt{2}$
   4. D.$-12 \sqrt{2}$
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   Answer: (A)

   $\left.\begin{array}{l}4 x-3 y=12 \alpha \\ 4 x+3 y=\frac{12}{\alpha}\end{array}\right\} \rightarrow 16 x^2-9 y^2=144$ $\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{16}=1 \rightarrow \text { Curve :S } \\ & T: y=m x \pm \sqrt{9 m^2-16} \\ & m=\frac{4 \sqrt{2}}{3} \\ & y=\frac{4 \sqrt{2} x}{3} \pm \sqrt{32-16} \\ & 3 y=4 \sqrt{2} x \pm 12 \\ & \text { as } q\gt 0 \\ & 3 y=4 \sqrt{2} x+12 \\ & p=-\frac{3}{\sqrt{2}} \& q=4 \\ & p q=-6 \sqrt{2}\end{aligned}$
2. Q2JEE Advanced Adv 2008 (Paper 1)
   Let $a$ and $b$ be non-zero and real numbers. Then, the equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ represents
   1. A.Four straight lines, when $c=0$ and $a, b$ are of the same sign
   2. B.Two straight lines and a circle, when $a=b$ and $c$ is of sign opposite to that of $a$
   3. C.Two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$
   4. D.A circle and an ellipse, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$
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   Answer: (B)

   Let $a$ and $b$ be non- zero real numbers. Therefore, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ implies either $$\begin{array}{rc} & x^2-5 x y+6 y^2=0 \\ \Rightarrow & (x-2 y)(x-3 y)=0 \\ \Rightarrow & x=2 y \text { and } x=3 y \end{array}$$ represent two straight lines passing through origin. Or $\quad a x^2+b y^2+c=0$ When $c=0$ and $a$ and $b$ are of same signs, then $$\begin{aligned} & a x^2+b y^2+c=0 \\ \Rightarrow & x=0 \text { and } y=0 \end{aligned}$$ which is a point specified as the origin. When $a=b$ and $c$ is of sign opposite to that of $a, a x^2+b y^2+c=0$ represents a circle. Hence, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ may represent two straight lines and a circle.