Let A(1, 2) be a point on the parabola y^2 = | vedclass

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(A) Let the coordinates of points $B$ and $C$ on the parabola $y^2 = 4x$ be $(t_1^2, 2t_1)$ and $(t_2^2, 2t_2)$ respectively.The line passing through

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is always right-angled

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(A) Let the coordinates of points $B$ and $C$ on the parabola $y^2 = 4x$ be $(t_1^2, 2t_1)$ and $(t_2^2, 2t_2)$ respectively.The line passing through $B(t_1^2, 2t_1)$ and $C(t_2^2, 2t_2)$ also passes through $P(5, -2)$.The slope of the line $BC$ is $m = \frac{2t_1 - 2t_2}{t_1^2 - t_2^2} = \frac{2}{t_1 + t_2}$.Since the line passes through $P(5, -2)$,the slope is also $m = \frac{2t_1 - (-2)}{t_1^2 - 5} = \frac{2(t_1 + 1)}{t_1^2 - 5}$.Equating the slopes: $\frac{2}{t_1 + t_2} = \frac{2(t_1 + 1)}{t_1^2 - 5}$ $\Rightarrow t_1^2 - 5 = (t_1 + t_2)(t_1 + 1) = t_1^2 + t_1 + t_1t_2 + t_2$.This simplifies to $t_1t_2 + t_1 + t_2 = -5$.The slopes of lines $AB$ and $AC$ are $m_{AB} = \frac{2t_1 - 2}{t_1^2 - 1} = \frac{2}{t_1 + 1}$ and $m_{AC} = \frac{2}{t_2 + 1}$.The product of the slopes is $m_{AB} \cdot m_{AC} = \frac{4}{(t_1 + 1)(t_2 + 1)} = \frac{4}{t_1t_2 + t_1 + t_2 + 1}$.Substituting $t_1t_2 + t_1 + t_2 = -5$,we get $m_{AB} \cdot m_{AC} = \frac{4}{-5 + 1} = \frac{4}{-4} = -1$.Since the product of the slopes is $-1$,the lines $AB$ and $AC$ are perpendicular,meaning $\angle BAC = 90^\circ$. Thus,$\Delta ABC$ is always right-angled.

Let $A(1, 2)$ be a point on the parabola $y^2 = 4x$. Let $B$ and $C$ be the points of intersection of this parabola with a variable line passing through the point $P(5, -2)$. Then the $\Delta ABC$ (if it exists):

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