Method 1: Direct Calculation

1. Find the vertices of the triangle. The triangle is formed by the intersection of the three lines:

   • L1: $(1 + p)x - py + p(1 + p) = 0$
   • L2: $(1 + q)x - qy + q(1 + q) = 0$
   • L3: $y = 0$

   Let's find the intersection points (vertices A, B, C).

   • Vertex C (Intersection of L1 and L3): Substitute $y=0$ into L1: $(1 + p)x - p(0) + p(1 + p) = 0$ $(1 + p)x = -p(1 + p)$ Assuming $p \neq -1$ (otherwise L1 becomes $y=0$, which is L3, and no triangle is formed), we get $x = -p$. So, vertex C is $(-p, 0)$.

   • Vertex B (Intersection of L2 and L3): Substitute $y=0$ into L2: $(1 + q)x - q(0) + q(1 + q) = 0$ $(1 + q)x = -q(1 + q)$ Assuming $q \neq -1$, we get $x = -q$. So, vertex B is $(-q, 0)$.

   • Vertex A (Intersection of L1 and L2): We solve the system of equations: $(1 + p)x - py = -p(1 + p) \quad \text{(1)}$ $(1 + q)x - qy = -q(1 + q) \quad \text{(2)}$ Multiply (1) by $q$ and (2) by $p$: $q(1 + p)x - pqy = -pq(1 + p)$ $p(1 + q)x - pqy = -pq(1 + q)$ Subtracting the second new equation from the first: $(q(1+p) - p(1+q))x = -pq(1+p) + pq(1+q)$ $(q + pq - p - pq)x = -pq - p^2q + pq + pq^2$ $(q-p)x = pq(q-p)$ Since $p \neq q$ (given), we can divide by $(q-p)$: $x = pq$ Substitute $x=pq$ into equation (1) to find $y$: $(1+p)(pq) - py = -p(1+p)$ pq + p^2q - py = -p - p^2$$ Assuming p \neq 0$, divide by$p$:$q + pq - y = -1 - p$$y = pq + p + q + 1 = (p+1)(q+1)$$ So, vertex A is $(pq, (1+p)(1+q))$.

2. Find the equations of the altitudes. Let the orthocentre be $H(h, k)$.

   • Altitude from A to BC: The side BC lies on the line $y=0$ (the x-axis). The altitude from A to BC is a vertical line passing through A. The equation of this altitude is $x = x_A$, which is $x = pq$. Since the orthocentre lies on this altitude, its x-coordinate must be $h = pq$.

   • Altitude from B to AC: The side AC is the line L1: $(1 + p)x - py + p(1 + p) = 0$. The slope of AC is $m_{AC} = -\frac{1+p}{-p} = \frac{1+p}{p}$. The slope of the altitude from B must be the negative reciprocal of $m_{AC}$, which is $m_{alt} = -\frac{p}{1+p}$. The altitude passes through vertex B $(-q, 0)$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 0 = -\frac{p}{1+p}(x - (-q))$ $y = -\frac{p}{1+p}(x+q)$

3. Find the coordinates of the orthocentre. The orthocentre $H(h, k)$ is the intersection of the altitudes. We already know $h = pq$. Substitute $x=h$ and $y=k$ into the equation of the second altitude: $k = -\frac{p}{1+p}(h+q)$ Now substitute $h = pq$: $k = -\frac{p}{1+p}(pq+q)$ $k = -\frac{p}{1+p}q(p+1)$ Since $p \neq -1$, we can cancel the $(p+1)$ terms: $k = -pq$

4. Determine the locus. The coordinates of the orthocentre are $(h, k) = (pq, -pq)$. We have the relations: $h = pq$ $k = -pq$ From these, we can eliminate the parameters $p$ and $q$ to find a relationship between $h$ and $k$: $k = -h$ Replacing $(h, k)$ with $(x, y)$ for the locus, we get: $y = -x \quad \text{or} \quad x+y=0$ This is the equation of a straight line.

Method 2: Using Properties of Parabola

1. The family of lines given by $(1 + p)x - py + p(1 + p) = 0$ can be written as a quadratic in the parameter $p$: $x + px - py + p + p^2 = 0 \implies p^2 + p(x-y+1) + x = 0$.
2. The envelope of this family of lines is found by setting the discriminant to zero: $B^2 - 4AC = 0$. $(x-y+1)^2 - 4(1)(x) = 0 \implies (x-y+1)^2 = 4x$. This is the equation of a parabola. The lines L1 and L2 are tangents to this parabola for parameters $p$ and $q$.
3. Let's check if the third line of the triangle, $y=0$, is also a tangent to this parabola. Substitute $y=0$ into the parabola's equation: $(x-0+1)^2 = 4x \implies (x+1)^2 = 4x \implies x^2+2x+1 = 4x \implies x^2-2x+1=0 \implies (x-1)^2=0$. This gives a repeated root $x=1$, which means $y=0$ is indeed a tangent to the parabola at the point $(1,0)$.
4. A well-known theorem in geometry states that the orthocentre of a triangle formed by three tangents to a parabola lies on the directrix of the parabola.
5. We need to find the directrix of the parabola $(x-y+1)^2 = 4x$. The equation is of the form $x^2-2xy+y^2-2x-2y+1=0$. The axis of this parabola is parallel to $x-y=0$. The directrix is perpendicular to the axis, so its equation will be of the form $x+y=c$. By rotating the coordinate axes by $45^\circ$ ($x = \frac{X-Y}{\sqrt{2}}, y = \frac{X+Y}{\sqrt{2}}$), the equation of the parabola transforms to $Y^2 = \sqrt{2}(X - \frac{\sqrt{2}}{4})$, whose directrix is $X=0$. Transforming back, $X = \frac{x+y}{\sqrt{2}}$, so the directrix is $\frac{x+y}{\sqrt{2}}=0$, which simplifies to $x+y=0$.
6. Therefore, the locus of the orthocentre is the straight line $x+y=0$.

Both methods lead to the same conclusion: the locus is a straight line. Comparing with the options:

A: a hyperbola - Incorrect. B: a parabola - Incorrect. C: an ellipse - Incorrect. D: a straight line - Correct.