If \( O \) is the vertex of the parabola \( x^2 = 4y \), \( Q \) is a point on the parabola. If \( C \) is the locus of the point which divides \( OQ \) in the ratio \( 2:3 \), then the equation of the chord of \( C \) which is bisected at the point \( (1,2) \) is:

Question

The area (in square units) of the region bounded by the parabola \(y^2 = 4(x - 2)\) and the line \(y = 2x - 8\) is:

Answer 1

To solve this problem, we need to determine the equation of the chord of the parabola \( x^2 = 4y \) that is bisected by the point \( (1, 2) \).

Given the vertex of the parabola \( x^2 = 4y \) is at \( O(0, 0) \).
Let \( Q(x_1, y_1) \) be a point on the parabola, which means \( x_1^2 = 4y_1 \).
The point that divides \( OQ \) in the ratio \( 2:3 \) can be found using the section formula: \(P\left(\frac{2 \cdot x_1 + 3 \cdot 0}{2 + 3}, \frac{2 \cdot y_1 + 3 \cdot 0}{2 + 3}\right) = \left(\frac{2x_1}{5}, \frac{2y_1}{5}\right)\)
The locus of point \( P \) as \( Q \) varies is given by substituting \( x_1^2 = 4y_1 \) into: \(\frac{4x^2}{25} = 4 \cdot \frac{y}{5}\), simplifying gives: \(x^2 = 4y\)
The chord with midpoint \( (1, 2) \) is found using the midpoint formula:
If the endpoints of the chord are \( (x_1, y_1) \) and \( (x_2, y_2) \), then: \(\frac{x_1 + x_2}{2} = 1\) and \(\frac{y_1 + y_2}{2} = 2\)
This gives two equations:
- \(x_1 + x_2 = 2\)
- \(y_1 + y_2 = 4\)
The equation of the chord is of the harmonic form for a parabola: \(T = S_1\) which translates to: \(xx_1 = y + y_1\) and \(xx_2 = y + y_2\), yielding the chord equation \(xx_0 = yy_0\)
Plug in \( x_0 = 1 \) and \( y_0 = 2 \) to solve \( x^2 - 2y - 2x + 4 = 0 \) for possible simplification.
Checking through options, the correct transformation from parabola form to chord results: \(5x - 4y + 3 = 0\)

Thus, the correct equation for the chord of the parabola bisected at point \( (1, 2) \) is \(5x - 4y + 3 = 0\).

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