Question:

Step 1: To find the locus of point $P$, we start by considering the equation of the circle $x^2 + y^2 = 1$ and a point $P$ with coordinates $(h,k)$. The combined equation of tangents from $P$ to the given circle can be derived using the point-circle tangents equation. This equation will help us establish a relationship between the coordinates of $P$ and the circle. Step 2: The equation of the tangents from point $P(h,k)$ to the circle $x^2 + y^2 - 1 = 0$ can be written as $y^2 - 2ky + k^2 + x^2 - 2hx + h^2 - 1 = 0$. However, to simplify the problem, we consider the condition that these tangents make an intercept of 2 units on the line $x = 1$. Setting $x = 1$ in the equation of the tangents, we get $(h-1)^2 + 2ky(h-1) = y^2(h^2-1)$, which simplifies to $y^2(h+1) - 2ky - (h-1) = 0$. Step 3: The given condition states that the tangents make an intercept of 2 units on the line $x = 1$. This implies that the distance between the points where the tangents intersect the line $x = 1$ is 2 units. Using the quadratic equation $y^2(h+1) - 2ky - (h-1) = 0$, we can find the roots $y_1$ and $y_2$ and then calculate the distance $AB = |y_1 - y_2| = 2$. Applying the quadratic formula and simplifying yields $4 = \frac{4k^2}{(h+1)^2} + \frac{4(h-1)}{h+1}$. Step 4: Solving the equation $4 = \frac{4k^2}{(h+1)^2} + \frac{4(h-1)}{h+1}$ gives us $k^2 = 2(h+1)$. This equation represents the relationship between the $x$ and $y$ coordinates of point $P$. To express the locus of $P$ in terms of $x$ and $y$, we substitute $h = x$ and $k = y$ into the equation $k^2 = 2(h+1)$, obtaining $y^2 = 2(x+1)$. Step 5: The equation $y^2 = 2(x+1)$ represents the locus of point $P$. This equation is in the form of a parabola, where $y^2$ is proportional to $x$. Therefore, the locus of $P$ is a parabola. Matching this result with the given options, we find that the correct answer is Option 1: parabola. Thus, the final answer is $\boxed{1}$.