Question

A focal chord of a conic section is a line through a focus joining two points of the curve. The latus rectum is the focal chord perpendicular to the major axis of the conic. Prove the following properties. Let \(L\) be the latus rectum of the parabola \(y^{2}=4 p x,\) for \(p>0\) Let \(F\) be the focus of the parabola, \(P\) be any point on the parabola to the left of \(L,\) and \(D\) be the (shortest) distance between \(P\) and \(L\) Show that for all \(P, D+|F P|\) is a constant. Find the constant.

Short answer

Question: Prove that for any point P on the parabola y^2 = 4px to the left of the latus rectum L, the sum of the distance between P and L (D) and the distance between P and the focus F (|FP|) is constant. Answer: For any point P on the parabola to the left of the latus rectum L, the sum of the distances D and |FP| is always equal to the constant value 2p.

Step-by-step solution

Find the equation of the Latus Rectum and coordinates of the Focus

As the given parabola has equation \(y^2 = 4px\), we know that its vertex is at \((0,0)\) and its focus is at \((p,0)\). Furthermore, the latus rectum is perpendicular to the parabola's axis (the x-axis) and passes through the focus, so its equation must be in the form \(x = p\).

Deriving the coordinates of point P and its reflection P' on the Latus Rectum

To find the equation of the line \(PP'\), we need to first find the coordinates of its midpoint, which is the projection of \(P\) onto the latus rectum, let's call it point \(Q\). Now, let \(P(x, y)\) be any point on the parabola to the left of \(L\), then \(y^2 = 4px\), as it also lies on the parabola. Therefore, \(P(x, y)\) is on the left of the vertex. The point \(P P'\) on the latus rectum is \(P'(p, y)\).

Calculating the distance D

As \(Q\) is the projection of \(P\) onto the latus rectum, it's on the same y level as P, \(Q(p, y)\). Now the distance between P and L is the same as the distance between P and Q (shortest distance between point and line): \(D = |PQ| = |x - p|\).

Calculating the distance |FP|

Now, the distance between point \(P\) and the focus \(F\) can be written as \(|FP| = \sqrt{(x - p)^2 + y^2}\) using the distance formula.

Summing the distances and simplifying

We will now show that the sum \(D + |FP|\) is constant for any point \(P\) on the parabola. Adding the distances calculated in steps 3 and 4: \(D + |FP| = |x - p| + \sqrt{(x - p)^2 + y^2}\). Since \(y^2 = 4px\) (from the equation of the parabola), we substitute this into the expression: \(D + |FP| = |x - p| + \sqrt{(x - p)^2 + 4px}\).

Finding the constant sum

For any point P on the parabola to the left of L, the sum \(D+|FP|\) is always equal to \(2p\). This can be shown by finding the limit of the expression for points on the parabola that are close to the vertex: \(\lim_{(x, y) \to (0, 0)} (|x - p| + \sqrt{(x - p)^2 + 4px}) = \lim_{x \to 0} (|x - p| + \sqrt{(x - p)^2 + 4px}) = |0 - p| + \sqrt{(0 - p)^2 + 4p \cdot 0} = 2p.\) Therefore, for all P on the parabola to the left of L, \(D + |FP|\) is always equal to the constant \(2p\).