Hyperbola
DAILY_CHALLENGE
Grade None
Question:
Let $S$ denote the locus of the point of intersection of the pair of lines
$$4x - 3y = 12\alpha ,$$
$$4\alpha x + 3\alpha y = 12 ,$$
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 $4x - \dfrac{3}{\sqrt{2}}y = 0$.
Then the value of $pq$ is
$-6\sqrt{2}$
$-3\sqrt{2}$
$-9\sqrt{2}$
$-12\sqrt{2}$
Step-by-Step Solution
Key Concept: Eliminating the parameter $\alpha$ to find the locus as a standard hyperbola, then using the slope form of the tangent to find its intercepts.
From the first line, we get $\alpha = \dfrac{4x - 3y}{12}$.
Substitute this into the second equation $\alpha(4x + 3y) = 12$:
$$\left(\dfrac{4x - 3y}{12}\right)(4x + 3y) = 12 \implies 16x^2 - 9y^2 = 144 \implies \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$$
This is a standard hyperbola with $a^2 = 9$ and $b^2 = 16$.
The tangent $T$ is parallel to the line $4x - \dfrac{3}{\sqrt{2}}y = 0 \implies y = \dfrac{4\sqrt{2}}{3}x$. Thus, the slope of $T$ is $m = \dfrac{4\sqrt{2}}{3}$.
The equation of a tangent with slope $m$ to the hyperbola is:
$$y = mx \pm \sqrt{a^2m^2 - b^2}$$
$$a^2m^2 - b^2 = 9\left(\dfrac{32}{9}\right) - 16 = 32 - 16 = 16$$
Thus, the equation of the tangent is:
$$y = \dfrac{4\sqrt{2}}{3}x \pm 4 \implies 4\sqrt{2}x - 3y \pm 12 = 0$$
Since the tangent passes through $(0, q)$ with $q > 0$, the $y$-intercept $q$ must be positive, which gives $q = 4$. This corresponds to the tangent line:
$$4\sqrt{2}x - 3y + 12 = 0$$
To find the $x$-intercept $p$, set $y=0$:
$$4\sqrt{2}p + 12 = 0 \implies p = -\dfrac{3}{\sqrt{2}}$$
Hence, the product is:
$$pq = \left(-\dfrac{3}{\sqrt{2}}\right)(4) = -6\sqrt{2}$$
Correct Answer: A