Let form a point `A(h, k)` chords of contact are drawn to the ellipse `x^(2)+2y^(2)=6` where all these chords touch the ellipse `x^(2)+4y^(2)=4`. Then, the perimeter (in units) of the locus of point A is

Options

A. `2pi`
B. `3pi`
C. `4pi`
D. `6pi`

Answer

To solve the problem, we need to find the perimeter of the locus of the point \( A(h, k) \) from which chords of contact are drawn to the ellipse \( x^2 + 2y^2 = 6 \) that touch the ellipse \( x^2 + 4y^2 = 4 \).

### Step-by-Step Solution:

1. **Identify the Ellipses**:
- The first ellipse is given by the equation \( x^2 + 2y^2 = 6 \).
- The second ellipse is given by the equation \( x^2 + 4y^2 = 4 \).

2. **Equation of Chord of Contact**:
- For a point \( A(h, k) \), the equation of the chord of contact to the first ellipse \( x^2 + 2y^2 = 6 \) is given by:
\[ \frac{hx}{6} + \frac{2ky}{6} = 1 \quad \text{(or)} \quad hx + 2ky = 6 \]

3. **Tangent to the Second Ellipse**:
- The equation of the tangent to the second ellipse \( x^2 + 4y^2 = 4 \) can be expressed as:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]
- The tangent line has the form:
\[ y = mx \pm \sqrt{4m^2 + 1} \]

4. **Finding the Slope**:
- The slope \( m \) of the chord of contact can be derived from the equation \( hx + 2ky = 6 \). Rearranging gives:
\[ y = -\frac{h}{2k}x + \frac{6}{2k} \]
- Thus, the slope \( m = -\frac{h}{2k} \).

5. **Setting Up the Condition**:
- The y-intercept of the chord of contact is \( \frac{6}{2k} \). Setting this equal to the y-intercept from the tangent line gives:
\[ \frac{6}{2k} = \pm \sqrt{4m^2 + 1} \]
- Substituting \( m = -\frac{h}{2k} \):
\[ \frac{6}{2k} = \pm \sqrt{4\left(-\frac{h}{2k}\right)^2 + 1} \]
- Squaring both sides leads to:
\[ \left(\frac{6}{2k}\right)^2 = 4\left(\frac{h^2}{4k^2}\right) + 1 \]
- Simplifying gives:
\[ \frac{36}{4k^2} = \frac{h^2}{k^2} + 1 \]

6. **Rearranging the Equation**:
- Multiply through by \( 4k^2 \):
\[ 36 = 4h^2 + 4k^2 \]
- Rearranging gives:
\[ 4h^2 + 4k^2 = 36 \quad \Rightarrow \quad h^2 + k^2 = 9 \]

7. **Finding the Locus**:
- The equation \( h^2 + k^2 = 9 \) represents a circle centered at the origin with a radius of \( 3 \).

8. **Calculating the Perimeter**:
- The perimeter \( P \) of a circle is given by the formula:
\[ P = 2\pi r \]
- Substituting \( r = 3 \):
\[ P = 2\pi \times 3 = 6\pi \]

### Final Answer:
The perimeter of the locus of point \( A \) is \( 6\pi \) units.