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 equation of the parabola
First, we need to rewrite the given equation of the parabola in a standard form. The given equation is:
\[
y^2 + 4y = \frac{4}{3}x - \frac{16}{3}
\]
We can rearrange it to:
\[
y^2 + 4y + \frac{16}{3} = \frac{4}{3}x
\]
Next, we complete the square for the \(y\) terms:
\[
(y + 2)^2 = \frac{4}{3}(x - 1)
\]
This shows that the parabola opens to the right with vertex at \((1, -2)\).
### Step 2: Identify the parameters of the parabola
In the standard form \( (y - k)^2 = 4a(x - h) \), we identify:
- Vertex \((h, k) = (1, -2)\)
- \(4a = \frac{4}{3}\) implies \(a = \frac{1}{3}\)
### Step 3: Parametric equations for points on the parabola
The parametric equations for points \(P\) and \(Q\) on the parabola can be given as:
\[
P(t_1) = \left(1 + \frac{1}{3}t_1^2, -2 + \frac{2}{3}t_1\right)
\]
\[
Q(t_2) = \left(1 + \frac{1}{3}t_2^2, -2 + \frac{2}{3}t_2\right)
\]
### Step 4: Condition for right angle at the vertex
For the chord \(PQ\) to subtend a right angle at the vertex \((1, -2)\), the product of the slopes of \(OP\) and \(OQ\) must equal \(-1\). The slopes are given by:
\[
\text{slope of } OP = \frac{-2 + \frac{2}{3}t_1 + 2}{1 + \frac{1}{3}t_1^2 - 1} = \frac{\frac{2}{3}t_1}{\frac{1}{3}t_1^2} = \frac{2}{t_1}
\]
\[
\text{slope of } OQ = \frac{-2 + \frac{2}{3}t_2 + 2}{1 + \frac{1}{3}t_2^2 - 1} = \frac{\frac{2}{3}t_2}{\frac{1}{3}t_2^2} = \frac{2}{t_2}
\]
Thus, we have:
\[
\frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \implies \frac{4}{t_1 t_2} = -1 \implies t_1 t_2 = -4
\]
### Step 5: Substitute \(t_1 t_2\) into the equation of the chord
The equation of the chord can be expressed as:
\[
y(t_1 + t_2) = 2x + 2a(t_1 t_2)
\]
Substituting \(t_1 t_2 = -4\):
\[
y(t_1 + t_2) = 2x - \frac{8}{3}
\]
### Step 6: Find the coordinates of the point through which the chord passes
To find the point through which the chord passes, we can set \(y = 0\):
\[
0(t_1 + t_2) = 2x - \frac{8}{3} \implies 2x = \frac{8}{3} \implies x = \frac{4}{3}
\]
Thus, the point is \(\left(\frac{4}{3}, 0\right)\).
### Step 7: Final coordinates of the point
Since the vertex of the parabola is \((1, -2)\), we can find the coordinates of the point where the chord passes through:
\[
x = 1 + \frac{4}{3} \implies x = \frac{7}{3}
\]
Thus, the coordinates of the point through which the chord passes are:
\[
\left(\frac{7}{3}, -2\right)
\]
### Conclusion
The final answer is the point \(\left(\frac{7}{3}, -2\right)\).
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