The locus of the foot of the perpendicular drawn from the centre on any tangent to the ellipse `x^2/25+y^2/16=1` is:

Options

A. `25x^2+16y^2=(x^2+y^2)^2`
B. `25x^2+16y^2=9`
C. `16x^2+25y^2=1`
D. `x^2+y^2=41`

Answer

To find the locus of the foot of the perpendicular drawn from the center of the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) to any tangent of the ellipse, we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse is in the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where:
- \( a^2 = 25 \) (thus \( a = 5 \))
- \( b^2 = 16 \) (thus \( b = 4 \))

### Step 2: Write the equation of the tangent
The equation of the tangent to the ellipse at a point \( (x_1, y_1) \) on the ellipse is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
For our ellipse, this becomes:
\[ \frac{xx_1}{25} + \frac{yy_1}{16} = 1 \]

### Step 3: Find the slope of the tangent
If we express the tangent in slope-intercept form, we can rearrange it:
\[ y = -\frac{16}{25} \frac{x_1}{y_1} x + \frac{16}{y_1} \]
The slope \( m_1 \) of the tangent line is:
\[ m_1 = -\frac{16}{25} \frac{x_1}{y_1} \]

### Step 4: Determine the foot of the perpendicular
Let \( (h, k) \) be the foot of the perpendicular from the center \( (0, 0) \) to the tangent line. The slope of the perpendicular \( m_2 \) is the negative reciprocal of \( m_1 \):
\[ m_2 = \frac{25}{16} \frac{y_1}{x_1} \]
Using the point-slope form of the line, we have:
\[ k - 0 = m_2(h - 0) \implies k = \frac{25}{16} \frac{y_1}{x_1} h \]

### Step 5: Use the distance formula
The distance \( d \) from the center \( (0, 0) \) to the tangent line can be calculated using the formula:
\[ d = \frac{|\frac{0 \cdot x_1}{25} + \frac{0 \cdot y_1}{16} - 1|}{\sqrt{\left(\frac{x_1}{25}\right)^2 + \left(\frac{y_1}{16}\right)^2}} = \frac{1}{\sqrt{\frac{x_1^2}{625} + \frac{y_1^2}{256}}} \]

### Step 6: Equate the distances
The distance from the center to the point \( (h, k) \) is:
\[ \sqrt{h^2 + k^2} \]
Setting the two distances equal gives:
\[ \sqrt{h^2 + k^2} = \frac{1}{\sqrt{\frac{x_1^2}{625} + \frac{y_1^2}{256}}} \]

### Step 7: Substitute \( k \) in terms of \( h \)
Substituting \( k = \frac{25}{16} \frac{y_1}{x_1} h \) into the distance equation:
\[ \sqrt{h^2 + \left(\frac{25}{16} \frac{y_1}{x_1} h\right)^2} = \frac{1}{\sqrt{\frac{x_1^2}{625} + \frac{y_1^2}{256}}} \]

### Step 8: Simplify and derive the locus
After simplifying, we can derive the relationship between \( h \) and \( k \) to find the locus. The final equation will represent the locus of the foot of the perpendicular.

The locus will be:
\[ \frac{25h^2}{25} + \frac{16k^2}{16} = 1 \]
This simplifies to:
\[ \frac{h^2}{25} + \frac{k^2}{16} = 1 \]
Thus, the locus of the foot of the perpendicular is an ellipse.