Question 3 (15 marks) Use a SEPARATE writing booklet. - HSC - SSCE Mathematics Extension 2 - Question 3 - 2002 - Paper 1

For $y^2 = f(x)$, this indicates that for every positive value of $f(x)$ there are two corresponding $y$ values: one positive and one negative. The graph should reflect the shape of $f(x)$ but mirrored above and below the x-axis. Ensure to correctly draw the points (1, 1) and (1, -1) as well as the curves approaching infinity in both directions.

To sketch $y = |f(x)|$, observe that any negative portion of the graph of $f(x)$ must be reflected upward. This means wherever $f(x)$ is negative, $|f(x)|$ will be positive. Include the points (1, 1) and (1, -1) appropriately adjusted. The result should show the negative section of the graph mirrored above the x-axis.

The function $y = \ln(f(x))$ is defined where $f(x) > 0$. The graph will exist only in areas where $f(x)$ is positive. Identify intercepts such as (2, 0) which corresponds to $f(2)=1$. As $f(x)$ approaches 0, $\ln(f(x))$ will approach $- \text{infinity}$. Sketch with respect to the behavior of $f(x)$ remaining positive.

To find the equation of the tangent at point $P(c p, \frac{c}{p})$, we use implicit differentiation on $xy = c^2$. First, differentiate to find the slope at $P$. The slope ($m$) can be calculated as $m = -\frac{q}{p}$. Then, we apply the point-slope form of the equation: [y - \frac{c}{p} = -\frac{q}{p}(x - cp)] Simplifying leads us to the desired tangent equation: [x + p^2y = 2cp.]

Using the coordinates of point $T$, we substitute the tangent equations into a system to find $T$. By equating two tangent lines via finding the intersection of the tangents at $P$ and $Q$, we can derive that the coordinates of point $T$ resolve to be [\left(\frac{2c pq}{p + q}, \frac{2c}{p + q}\right).]

To show that the locus of $T$ is a hyperbola, we can substitute the coordinates we've found back into the hyperbolic equation form derived from $xy = c^2$. Rewrite the variables $x$ and $y$ in terms of $p$ and $q$, and simplify accordingly. This will shape a hyperbola indicating that as points $P$ and $Q$ move, so does $T$ along a hyperbola. The eccentricity can also be computed from given parameters, noting that it relates to the distance away from the center of the hyperbola curve.