Let from a point A (h,k) chord of contacts are drawn to the ellipse `x^2+2y^2=6` such that all these chords touch the ellipse `x^2+4y^2=4,` then locus of the point A is

Options

A. `4x^2+9y^2=36`
B. `x^2+y^2=4`
C. `x^2-y^2=9`
D. `x^2+y^2=9`

Answer

To find the locus of the point \( A(h, k) \) from which chords of contact are drawn to the ellipse \( x^2 + 2y^2 = 6 \) such that all these chords touch the ellipse \( x^2 + 4y^2 = 4 \), we can follow these steps:

### Step 1: Write the equations of the ellipses
The first ellipse can be rewritten as:
\[ \frac{x^2}{6} + \frac{y^2}{3} = 1 \]
The second ellipse can be rewritten as:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]

### Step 2: Equation of the chord of contact
The equation of the chord of contact from a point \( (h, k) \) to the first ellipse \( x^2 + 2y^2 = 6 \) is given by:
\[ \frac{hx}{6} + \frac{ky}{3} = 1 \]

### Step 3: Parametric coordinates of the second ellipse
The parametric coordinates of any point on the second ellipse \( x^2 + 4y^2 = 4 \) can be expressed as:
\[ (2 \cos \theta, \sin \theta) \]

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

### Step 5: Equate the two equations
Since the chord of contact from point \( A(h, k) \) must touch the second ellipse, we equate the two equations:
\[ \frac{h}{6} x + \frac{k}{3} y = 1 \]
and
\[ \frac{\cos \theta}{2} x + \sin \theta y = 1 \]

### Step 6: Compare coefficients
From the equations, we can compare coefficients:
1. For \( x \): \( \frac{h}{6} = \frac{\cos \theta}{2} \)
2. For \( y \): \( \frac{k}{3} = \sin \theta \)

### Step 7: Express \( \cos \theta \) and \( \sin \theta \)
From the comparisons, we have:
\[ \cos \theta = \frac{h}{3} \quad \text{and} \quad \sin \theta = \frac{k}{3} \]

### Step 8: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \left(\frac{k}{3}\right)^2 + \left(\frac{h}{3}\right)^2 = 1 \]
This simplifies to:
\[ \frac{k^2}{9} + \frac{h^2}{9} = 1 \]
Multiplying through by 9 gives:
\[ k^2 + h^2 = 9 \]

### Step 9: Write the locus
Replacing \( h \) and \( k \) with \( x \) and \( y \):
\[ x^2 + y^2 = 9 \]

### Conclusion
Thus, the locus of the point \( A(h, k) \) is:
\[ \boxed{x^2 + y^2 = 9} \]