Let the point P of the focal chord PQ of the parabola y^2=16 x be (1,-4).

Question

Let the point P of the focal chord PQ of the parabola $y^2=16 x$ be $(1,-4)$. If the focus of the parabola divides the chord $P Q$ in the ratio $m: n, \operatorname{gcd}(m, n)=1$, then $m^2+n^2$ is equal to :

Accepted Answer

Given the parabola $ y^2 = 16x $, the point $ P $ on the focal chord $ PQ $ is $ (1, -4) $. We need to determine how the focus divides the chord.

First, express the coordinates for the point $ P $ using the parametric form of the parabola. The general parametric equations for the parabola are:

$ P(a t^2, 2 a t) \equiv P(4t^2, 8t) \equiv (1, -4) $

The equation $ 8t = -4 $ leads to solving for $ t $:

$ t = -\frac{1}{2} $

Now, find the coordinates of point $ Q $ using the relation $ t_1 \cdot t_2 = -1 $. Therefore:

$ Q\left(\frac{a}{t^2}, \frac{-2a}{t}\right) $

Given the focus $ S $ of the parabola is at $ (4, 0) $, we can find the lengths of $ PS $ and $ QS $:

The length $ PS $ is calculated as:

$ PS = a + a t^2 $

And $ QS $ is given by:

$ QS = a + \frac{a}{t^2} = \frac{a t^2 + a}{t^2} $

Since the focus divides the chord in a specific ratio $ m:n $, and $\gcd(m, n) = 1$, we find:

$ \frac{PS}{QS} = t^2 = \frac{1}{4} = \frac{m}{n} $

Thus, the squares of $ m $ and $ n $ sum up to:

$ m^2 + n^2 = 17 $

Solutions · 1