Question

Prove that portion of tangent to a parabola between point of contact and directrix always subtends a right angle at its focus.

Answer 1

The statement is a theorem in conic sections that needs to be proven. The proof involves setting up the standard equation of a parabola, defining a point of contact, finding the equation of the tangent at that point, determining the intersection of the tangent with the directrix, and then using vector dot products to show that the angle formed by the focus and these two points (point of contact and intersection with directrix) is 90 degrees.

Answer 2

Consider the standard equation of a parabola: $y^2 = 4ax$. The focus is $S(a, 0)$ and the directrix is $x = -a$. Let $P(at^2, 2at)$ be a point on the parabola. The tangent at $P$ is $yt = x + at^2$. Let $Q$ be the intersection of the tangent with the directrix ($x=-a$). Substituting $x=-a$, we get $y_Q t = -a + at^2$, so $y_Q = \frac{a(t^2-1)}{t}$. Thus, $Q = \left(-a, \frac{a(t^2-1)}{t}\right)$. We need to show that $\angle PSQ = 90^\circ$. This is equivalent to showing $\vec{SP} \cdot \vec{SQ} = 0$. $\vec{SP} = P - S = (at^2 - a, 2at) = (a(t^2-1), 2at)$. $\vec{SQ} = Q - S = \left(-a - a, \frac{a(t^2-1)}{t}\right) = \left(-2a, \frac{a(t^2-1)}{t}\right)$. $\vec{SP} \cdot \vec{SQ} = (a(t^2-1))(-2a) + (2at)\left(\frac{a(t^2-1)}{t}\right) = -2a^2(t^2-1) + 2a^2(t^2-1) = 0$. Since the dot product is zero, $\vec{SP}$ and $\vec{SQ}$ are orthogonal, hence $\angle PSQ = 90^\circ$.