The equation of the locus of the middle point of the portion of the tangent to the ellipse `x^/16+y^2/9=1` included between the co-ordinate axes is the curve

Options

A. `9x^(2)+16y^(2)=4x^(2)y^(2)`
B. `16x^(2)+9y^(2)=4x^(2)y^(2)`
C. `3x^(2)+4y^(2)=4x^(2)y^(2)`
D. `9x^(2)+16y^(2)=x^(2)y^(2)`

Answer

To find the equation of the locus of the midpoint of the portion of the tangent to the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) included between the coordinate axes, we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse is in the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, we have:
- \( a^2 = 16 \) which gives \( a = 4 \)
- \( b^2 = 9 \) which gives \( b = 3 \)

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

### Step 3: Find the intercepts of the tangent with the axes
To find the x-intercept (let's call it point A), set \( y = 0 \):
\[ \frac{x \cos \theta}{4} = 1 \implies x = \frac{4}{\cos \theta} \]
Thus, the coordinates of point A are \( \left(\frac{4}{\cos \theta}, 0\right) \).

To find the y-intercept (let's call it point B), set \( x = 0 \):
\[ \frac{y \sin \theta}{3} = 1 \implies y = \frac{3}{\sin \theta} \]
Thus, the coordinates of point B are \( \left(0, \frac{3}{\sin \theta}\right) \).

### Step 4: Calculate the midpoint of segment AB
The midpoint \( P \) of segment AB has coordinates:
\[ P\left(\frac{\frac{4}{\cos \theta} + 0}{2}, \frac{0 + \frac{3}{\sin \theta}}{2}\right) = \left(\frac{2}{\cos \theta}, \frac{3}{2 \sin \theta}\right) \]

### Step 5: Express \( \sin \theta \) and \( \cos \theta \) in terms of \( x \) and \( y \)
Let \( x = \frac{2}{\cos \theta} \) and \( y = \frac{3}{2 \sin \theta} \).

From these equations, we can express \( \cos \theta \) and \( \sin \theta \):
\[ \cos \theta = \frac{2}{x} \quad \text{and} \quad \sin \theta = \frac{3}{2y} \]

### Step 6: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \left(\frac{3}{2y}\right)^2 + \left(\frac{2}{x}\right)^2 = 1 \]
This simplifies to:
\[ \frac{9}{4y^2} + \frac{4}{x^2} = 1 \]

### Step 7: Clear the fractions
Multiplying through by \( 4y^2x^2 \) gives:
\[ 9x^2 + 16y^2 = 4y^2x^2 \]

### Final Equation
Rearranging gives us the locus of the midpoint:
\[ 9x^2 + 16y^2 - 4y^2x^2 = 0 \]

### Conclusion
The equation of the locus of the midpoint of the portion of the tangent to the ellipse included between the coordinate axes is:
\[ 9x^2 + 16y^2 = 4x^2y^2 \]