JEE MAIN - Mathematics (2025 - 2nd April Evening Shift - No. 4)
Explanation
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 $
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