The focal chord of the parabola `y^(2)=32x` touches the ellipse `(x^(2))/(4^(2))+(y^(2))/(2^(2))=1` in the first quadrant at the point

Options

A. `(2, sqrt3)`
B. `((sqrt3)/(2), (3)/(2))`
C. `((4)/(sqrt2),(2)/(sqrt2))`
D. `((2)/(sqrt2),(4)/(sqrt2))`

Answer

To solve the problem step by step, we need to find the point where the focal chord of the parabola \(y^2 = 32x\) touches the ellipse \(\frac{x^2}{4^2} + \frac{y^2}{2^2} = 1\) in the first quadrant.

### Step 1: Identify the properties of the parabola and ellipse
The given parabola is \(y^2 = 32x\). This can be rewritten in standard form as:
\[ x = \frac{y^2}{32} \]
The focus of this parabola is at the point \((8, 0)\) since the standard form of a parabola \(y^2 = 4ax\) gives \(a = 8\).

The given ellipse is \(\frac{x^2}{16} + \frac{y^2}{4} = 1\), which has semi-major axis \(a = 4\) and semi-minor axis \(b = 2\).

### Step 2: Parametrize the ellipse
The points on the ellipse can be expressed in parametric form as:
\[ (x, y) = (4 \cos \theta, 2 \sin \theta) \]
where \(\theta\) is the parameter.

### Step 3: Write the equation of the tangent to the ellipse
The equation of the tangent to the ellipse at the point \((4 \cos \theta, 2 \sin \theta)\) is given by:
\[ \frac{x \cos \theta}{4} + \frac{y \sin \theta}{2} = 1 \]

### Step 4: Substitute the focus of the parabola into the tangent equation
Since the tangent touches the parabola at the focus \((8, 0)\), we substitute \(x = 8\) and \(y = 0\) into the tangent equation:
\[ \frac{8 \cos \theta}{4} + \frac{0 \cdot \sin \theta}{2} = 1 \]
This simplifies to:
\[ 2 \cos \theta = 1 \implies \cos \theta = \frac{1}{2} \]

### Step 5: Find the value of \(\theta\)
The value of \(\theta\) that satisfies \(\cos \theta = \frac{1}{2}\) is:
\[ \theta = \frac{\pi}{3} \]

### Step 6: Calculate the coordinates of the point on the ellipse
Substituting \(\theta = \frac{\pi}{3}\) into the parametric equations of the ellipse:
\[ x = 4 \cos \left(\frac{\pi}{3}\right) = 4 \cdot \frac{1}{2} = 2 \]
\[ y = 2 \sin \left(\frac{\pi}{3}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]
Thus, the point where the focal chord of the parabola touches the ellipse in the first quadrant is:
\[ (2, \sqrt{3}) \]

### Final Answer
The point is \((2, \sqrt{3})\).