The locus of the feet of perpendiculars drawn from the centre of the ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1` on any tangent to it is .......... .

Answer

To find the locus of the feet of the perpendiculars drawn from the center of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) on any tangent to it, we can follow these steps:

### Step 1: Identify the center of the ellipse
The center of the ellipse given by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is at the point \((0, 0)\).

**Hint:** Remember that the center of the ellipse is always at the origin for the standard form of the ellipse.

### Step 2: Write the equation of the tangent
The equation of a tangent to the ellipse at a point \((x_1, y_1)\) on the ellipse can be expressed as:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
However, we can also express it in the slope-intercept form as:
\[ y = mx + \sqrt{a^2m^2 + b^2} \]
where \(m\) is the slope of the tangent.

**Hint:** The slope \(m\) can vary for different tangents, and the term \(\sqrt{a^2m^2 + b^2}\) represents the y-intercept.

### Step 3: Find the equation of the perpendicular from the center
The perpendicular from the center \((0, 0)\) to the tangent line can be expressed as:
\[ y = -\frac{1}{m}x \]
This line passes through the origin and has a slope of \(-\frac{1}{m}\).

**Hint:** The slope of the perpendicular line is the negative reciprocal of the slope of the tangent line.

### Step 4: Determine the foot of the perpendicular
Let the foot of the perpendicular be the point \((h, k)\). This point lies on both the tangent line and the perpendicular line. Thus, we can set up the equations:
1. From the tangent:
\[ k = mh + \sqrt{a^2m^2 + b^2} \]
2. From the perpendicular:
\[ k = -\frac{1}{m}h \]

**Hint:** You will need to solve these two equations simultaneously to find \(h\) and \(k\).

### Step 5: Substitute and simplify
Substituting \(k\) from the second equation into the first equation gives:
\[ -\frac{1}{m}h = mh + \sqrt{a^2m^2 + b^2} \]
Rearranging this leads to:
\[ -\frac{1}{m}h - mh = \sqrt{a^2m^2 + b^2} \]
Multiplying through by \(m\) to eliminate the fraction:
\[ -h - m^2h = m\sqrt{a^2m^2 + b^2} \]
Factoring out \(h\):
\[ h(-1 - m^2) = m\sqrt{a^2m^2 + b^2} \]
Thus,
\[ h = -\frac{m\sqrt{a^2m^2 + b^2}}{1 + m^2} \]

**Hint:** This expression for \(h\) will help you find the relationship between \(h\) and \(k\).

### Step 6: Express \(k\) in terms of \(h\)
Now substitute \(h\) back into the equation for \(k\):
\[ k = -\frac{1}{m}\left(-\frac{m\sqrt{a^2m^2 + b^2}}{1 + m^2}\right) = \frac{\sqrt{a^2m^2 + b^2}}{1 + m^2} \]

**Hint:** This will give you a parametric representation of the locus.

### Step 7: Eliminate \(m\)
To find the locus, we need to eliminate \(m\) from the equations for \(h\) and \(k\). Using the relationships derived, we can find:
\[ \frac{h^2}{a^2} + \frac{k^2}{b^2} = 1 \]

**Hint:** This is the standard form of the ellipse, which indicates that the locus of the feet of the perpendiculars is another ellipse.

### Final Result
Thus, the locus of the feet of the perpendiculars drawn from the center of the ellipse on any tangent to it is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

**Hint:** This confirms that the locus is indeed another ellipse, similar to the original one.