Chord of the parabola `y^2+4y=(4)/(3)x-(16)/(3)` which subtend right angle at the vertex pass through:

Options

A. `((7)/(3),-2)`
B. `((1)/(3),0)`
C. `((4)/(3),0)`
D. `(0,(4)/(3))`

Answer

To solve the problem of finding the chord of the parabola \( y^2 + 4y = \frac{4}{3}x - \frac{16}{3} \) that subtends a right angle at the vertex, we will follow these steps:

### Step 1: Rewrite the parabola in standard form
We start with the equation of the parabola:
\[ y^2 + 4y = \frac{4}{3}x - \frac{16}{3} \]
To rewrite it in standard form, we complete the square for the \( y \) terms:
\[ y^2 + 4y + 4 = \frac{4}{3}x - \frac{16}{3} + 4 \]
This simplifies to:
\[ (y + 2)^2 = \frac{4}{3}(x - 1) \]
This is now in the standard form of a parabola:
\[ (y - k)^2 = 4a(x - h) \]
where \( (h, k) = (1, -2) \) and \( 4a = \frac{4}{3} \).

### Step 2: Identify parameters
From the equation \( 4a = \frac{4}{3} \), we find:
\[ a = \frac{1}{3} \]
The vertex of the parabola is at \( (1, -2) \).

### Step 3: Parametric form of points on the parabola
Let the points on the parabola be represented parametrically as:
\[ P(t_1) = (at_1^2 + h, 2at_1 + k) = \left(\frac{1}{3}t_1^2 + 1, \frac{2}{3}t_1 - 2\right) \]
\[ Q(t_2) = (at_2^2 + h, 2at_2 + k) = \left(\frac{1}{3}t_2^2 + 1, \frac{2}{3}t_2 - 2\right) \]

### Step 4: Equation of the chord
The equation of the chord joining points \( P \) and \( Q \) is given by:
\[ y(t_1 + t_2) = 2x + 2a t_1 t_2 \]
Substituting \( a = \frac{1}{3} \):
\[ y(t_1 + t_2) = 2x + \frac{2}{3} t_1 t_2 \]

### Step 5: Condition for right angle at the vertex
For the chord to subtend a right angle at the vertex, the product of the slopes of the lines from the vertex to the points \( P \) and \( Q \) must equal -1:
\[ \left(\frac{2a t_1}{at_1^2}\right) \left(\frac{2a t_2}{at_2^2}\right) = -1 \]
This simplifies to:
\[ \frac{4}{3} \cdot \frac{t_1 t_2}{t_1^2 t_2^2} = -1 \]
Thus:
\[ t_1 t_2 = -4 \]

### Step 6: Substitute \( t_1 t_2 \) into the chord equation
Substituting \( t_1 t_2 = -4 \) into the chord equation:
\[ y(t_1 + t_2) = 2x - \frac{8}{3} \]

### Step 7: Find the coordinates of the intersection
The chord passes through the vertex \( (1, -2) \):
\[ -2(t_1 + t_2) = 2(1) - \frac{8}{3} \]
This leads to:
\[ -2(t_1 + t_2) = 2 - \frac{8}{3} = \frac{6 - 8}{3} = -\frac{2}{3} \]
Thus:
\[ t_1 + t_2 = \frac{1}{3} \]

### Step 8: Solve for \( t_1 \) and \( t_2 \)
Now we have a system of equations:
1. \( t_1 + t_2 = \frac{1}{3} \)
2. \( t_1 t_2 = -4 \)

Let \( t_1 \) and \( t_2 \) be the roots of the quadratic equation:
\[ x^2 - (t_1 + t_2)x + t_1 t_2 = 0 \]
This gives:
\[ x^2 - \frac{1}{3}x - 4 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{\frac{1}{3} \pm \sqrt{\left(\frac{1}{3}\right)^2 + 16}}{2} \]
Calculating the discriminant:
\[ \left(\frac{1}{3}\right)^2 + 16 = \frac{1}{9} + \frac{144}{9} = \frac{145}{9} \]
Thus:
\[ x = \frac{\frac{1}{3} \pm \frac{\sqrt{145}}{3}}{2} = \frac{1 \pm \sqrt{145}}{6} \]

### Step 10: Find the coordinates of the points
Substituting back to find the coordinates of the points \( P \) and \( Q \):
Using \( t_1 \) and \( t_2 \) in the parametric equations, we find the coordinates of the points through which the chord passes.

### Final Answer
The points through which the chord passes are:
\[ \left(\frac{7}{3}, -2\right) \]