The locus of poles of tangents to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` with respect to concentric ellipse `(x^(2))/(alpha^(2))+(y^(2))/(beta^(2))=1` is

Options

A. `a^2c^4x^2-b^2d^4y^2=1`
B. `a^2c^4x^2+b^2d^4y^2=1`
C. `2a^2c^4x^2-b^2d^4y^2=1`
D. `2a^2c^4x^2-b^2d^4y^2=2`

Answer

To find the locus of the poles of tangents to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with respect to the concentric ellipse \(\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1\), we can follow these steps:

### Step 1: Understand the concept of the pole and polar
The pole of a point with respect to a conic is a point from which tangents drawn to the conic will touch it. The equation of the polar line of a point \((h, k)\) with respect to the ellipse \(\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1\) is given by:
\[ \frac{xh}{\alpha^2} + \frac{yk}{\beta^2} = 1 \]

### Step 2: Write the equation of the tangent line
We can express the polar equation in the slope-intercept form \(y = mx + c\). Rearranging the polar equation gives:
\[ y = -\frac{\beta^2 h}{\alpha^2 k} x + \frac{\beta^2}{k} \]
This represents the tangent line to the ellipse.

### Step 3: Relate the tangent line to the ellipse
The tangent line must touch the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). The general equation of the tangent to this ellipse can be expressed as:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
For the tangents to be identical, the constant term \(c\) from the polar equation must equal the constant term from the tangent equation.

### Step 4: Equate the constant terms
From the tangent equation, we can equate the constant terms:
\[ \frac{\beta^2}{k} = \sqrt{a^2 m^2 + b^2} \]
Squaring both sides gives:
\[ \frac{\beta^4}{k^2} = a^2 m^2 + b^2 \]

### Step 5: Substitute for \(m\)
From the polar equation, we can express \(m\) in terms of \(h\) and \(k\):
\[ m = -\frac{\beta^2 h}{\alpha^2 k} \]
Substituting this into the equation derived in Step 4 gives:
\[ \frac{\beta^4}{k^2} = a^2 \left(-\frac{\beta^2 h}{\alpha^2 k}\right)^2 + b^2 \]
This simplifies to:
\[ \frac{\beta^4}{k^2} = \frac{a^2 \beta^4 h^2}{\alpha^4 k^2} + b^2 \]

### Step 6: Rearranging the equation
Multiplying through by \(k^2\) to eliminate the denominator gives:
\[ \beta^4 = a^2 \beta^4 \frac{h^2}{\alpha^4} + b^2 k^2 \]
Rearranging leads to:
\[ \frac{b^2 k^2}{\beta^4} + \frac{a^2 h^2}{\alpha^4} = 1 \]

### Step 7: Final locus equation
Substituting \(h\) and \(k\) with \(x\) and \(y\) respectively, we arrive at the locus equation:
\[ \frac{b^2 x^2}{\beta^4} + \frac{a^2 y^2}{\alpha^4} = 1 \]

### Conclusion
Thus, the locus of the poles of the tangents to the ellipse with respect to the concentric ellipse is given by:
\[ \frac{b^2 x^2}{\beta^4} + \frac{a^2 y^2}{\alpha^4} = 1 \]