JEE Advanced · Mathematics · 10. Pair of Lines
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 \((0, q), q>0\), and parallel to the line \(4 x-\frac{3}{\sqrt{2}} y=0\).
Then the value of \(p q\) is
- A \(-6 \sqrt{2}\)
- B \(-3 \sqrt{2}\)
- C \(-9 \sqrt{2}\)
- D \(-12 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(-6 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\left.\begin{array}{l}4 x-3 y=12 \alpha \\ 4 x+3 y=\frac{12}{\alpha}\end{array}\right\} \rightarrow 16 x^2-9 y^2=144\)
\(\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{16}=1 \rightarrow \text { Curve :S } \\ & T: y=m x \pm \sqrt{9 m^2-16} \\ & m=\frac{4 \sqrt{2}}{3} \\ & y=\frac{4 \sqrt{2} x}{3} \pm \sqrt{32-16} \\ & 3 y=4 \sqrt{2} x \pm 12 \\ & \text { as } q>0 \\ & 3 y=4 \sqrt{2} x+12 \\ & p=-\frac{3}{\sqrt{2}} \& q=4 \\ & p q=-6 \sqrt{2}\end{aligned}\)
\(\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{16}=1 \rightarrow \text { Curve :S } \\ & T: y=m x \pm \sqrt{9 m^2-16} \\ & m=\frac{4 \sqrt{2}}{3} \\ & y=\frac{4 \sqrt{2} x}{3} \pm \sqrt{32-16} \\ & 3 y=4 \sqrt{2} x \pm 12 \\ & \text { as } q>0 \\ & 3 y=4 \sqrt{2} x+12 \\ & p=-\frac{3}{\sqrt{2}} \& q=4 \\ & p q=-6 \sqrt{2}\end{aligned}\)
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