Parabola
Parabola
nta_abhyas_2025
Grade 11

Question:

Chord joining two distinct points $P(c_0, \frac{c_0}{2})$ and $Q(c_0, -\frac{c_0}{2})$ (both are variable points) on the parabola $y^2 = 16x$ always passes through a fixed point $(\alpha, \beta)$. Then, which of the following statements is correct?
$\alpha + \beta = 2$
$\alpha - \beta = 4$
$|\alpha| + |\beta| = 8$
$|\alpha| = |\beta|$

Step-by-Step Solution

Key Concept: For a parabola $y^2 = 4ax$, a focal chord satisfies the property $t_1 t_2 = -1$ where the endpoints are parameterized as $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$.
Given $P = (a, 4b)$, $Q = (-, \frac{1b}{c})$, and $O = (-, 8t_2)$. For a focal chord, if $P = (at_1^2, 2at_1)$ and $Q = (at_2^2, 2at_2)$, then $t_1 t_2 = -1$. With the given coordinates, we have $t_1 = \frac{2}{a}$ and $t_2 = -\frac{2}{a}$, so $t_1 t_2 = -1$. Chord $PQ$ passes through the focus, confirming it is a focal chord. The point $(a, \beta) = (4, 0)$ lies on the focal chord.
Correct Answer: 3

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