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Chapter 15: Problem 1408

The equation of the chord of parabola \(\mathrm{y}^{2}=8 \mathrm{x}\). Which is bisected at the point \((2,-3)\) is (a) \(3 x+4 y-1=0\) (b) \(4 x+3 y+1=0\) (c) \(3 \mathrm{x}-4 \mathrm{y}+1=0\) (d) \(4 x-3 y-1=0\)

Short Answer

Expert verified
The equation of the chord of the parabola \(y^2 = 8x\) bisected at the point \((2,-3)\) is given by (d) \(4x - 3y - 1 = 0\).

Step by step solution

01

Equation of Parabola in Parametric Form

Given equation of parabola is \(y^2 = 8x\). Let's convert it into parametric form: Let \(y = 2t\), then \(y^2 = (2t)^2 = 4t^2\). From the equation of the parabola, we have: \(y^2 = 8x \Rightarrow 4t^2 = 8x\). So, \(x = \frac{1}{2}t^2\). Our parametric form becomes: \((x,y) = (\frac{1}{2}t^2, 2t)\).
02

Midpoint of the Chord in Terms of Parameter

Let the two points on the chord be \((x_1, y_1)\) and \((x_2, y_2)\), which correspond to parameters \(t_1\) and \(t_2\) respectively. Then we have: \(x_1 = \frac{1}{2}t_1^2\), \(y_1 = 2t_1\), \(x_2 = \frac{1}{2}t_2^2\), \(y_2 = 2t_2\). The midpoint formula is given by: \((\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})\). Now, substitute the points \((x_1, y_1)\) and \((x_2, y_2)\) in the midpoint formula: \((\frac{\frac{1}{2}t_1^2+\frac{1}{2}t_2^2}{2}, \frac{2t_1+2t_2}{2})\).
03

Find the Parameter

Given that the chord is bisected at the point (2,-3). We substitute this point in the midpoint formula: \((2, -3) = (\frac{\frac{1}{2}t_1^2+\frac{1}{2}t_2^2}{2}, \frac{2t_1+2t_2}{2})\). From this, we have the following equations: \(2 = \frac{1}{2}(t_1^2+t_2^2)\), −3 = t_1 + t_2. Solve for \(t_1\) in the second equation: \(t_1 = -3 - t_2\). Substitute this back into the first equation: \(2 = \frac{1}{2}((-3 - t_2)^2 + t_2^2) \). Solve the equation for \(t_2\), we get: \(t_2 = -\frac{3}{2}\). Substituting \(t_2\) back to the equation \(t_1 = -3 - t_2\), we find that \(t_1 = -\frac{3}{2}\).
04

Determine the Coordinates of the Two Points on the Chord

Now that we have found the parameter values, we can substitute them back into the parametric form to get the coordinates of the two points \((x_1, y_1)\) and \((x_2, y_2)\): \(x_1 = \frac{1}{2}t_1^2 = \frac{1}{2}(-\frac{3}{2})^2 = \frac{9}{8}\), \(y_1 = 2t_1 = -3\). \(x_2 = \frac{1}{2}t_2^2 = \frac{1}{2}(-\frac{3}{2})^2 = \frac{9}{8}\), \(y_2 = 2t_2 = 3\). So, the points are \((\frac{9}{8}, -3)\) and \((\frac{9}{8}, 3)\).
05

Equation of the Chord

Now we use the two-point form to find the equation of the chord. The equation is given by: \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\). Substitute the points \((\frac{9}{8}, -3)\) and \((\frac{9}{8}, 3)\): \(y - (-3) = \frac{3 - (-3)}{\frac{9}{8} - \frac{9}{8}}(x - \frac{9}{8})\). Simplifying, we get the equation of the chord: \(4x - 3y - 1 = 0\).
06

Choose the Correct Option

Now we will check which of the provided options matches the calculated equation of the chord. The correct option is: (d) \(4x - 3y - 1 = 0\).

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