The locus of the poles of the tangents to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` w.r.t. the circle `x^2 + y^2 = a^2` is:

Options

A. parabola
B. ellipse
C. hyperbola
D. circle

Answer

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

### Step 1: Write the equation of the tangent to the ellipse
The equation of the tangent to the ellipse in slope-intercept form is given by:
\[ y = mx + \sqrt{a^2 m^2 + b^2} \]
where \(m\) is the slope of the tangent.

### Step 2: Identify the point on the locus of poles
Let \(P(H, K)\) be the point on the locus of the poles. The equation of the circle is \(x^2 + y^2 = a^2\). The equation of the pole with respect to the circle at point \(P(H, K)\) is given by:
\[ Hx + Ky = a^2 \]

### Step 3: Set up the equation for the tangents
The tangents to the ellipse must also satisfy the equation of the circle. Therefore, we can equate the tangent line and the pole equation:
\[ Hx + Ky = a^2 \]
Substituting \(y\) from the tangent equation into the pole equation gives:
\[ H x + K(mx + \sqrt{a^2 m^2 + b^2}) = a^2 \]

### Step 4: Rearranging the equation
Rearranging this gives:
\[ H x + K mx + K \sqrt{a^2 m^2 + b^2} = a^2 \]
This can be simplified to:
\[ (H + Km)x + K \sqrt{a^2 m^2 + b^2} = a^2 \]

### Step 5: Compare coefficients
For this equation to hold for all \(x\), the coefficient of \(x\) must be zero, leading to:
\[ H + Km = 0 \quad \Rightarrow \quad m = -\frac{H}{K} \]

### Step 6: Substitute \(m\) back into the equation
Now substitute \(m\) back into the equation:
\[ K \sqrt{a^2 m^2 + b^2} = a^2 \]
Substituting \(m = -\frac{H}{K}\) gives:
\[ K \sqrt{a^2 \left(-\frac{H}{K}\right)^2 + b^2} = a^2 \]
Squaring both sides leads to:
\[ K^2 \left(\frac{a^2 H^2}{K^2} + b^2\right) = a^4 \]
This simplifies to:
\[ a^2 H^2 + K^2 b^2 = a^4 \]

### Step 7: Rearranging to find the locus
Dividing through by \(a^4\) gives:
\[ \frac{H^2}{a^2} + \frac{K^2}{\frac{a^4}{b^2}} = 1 \]
This represents the equation of an ellipse.

### Conclusion
Thus, the locus of the poles of the tangents to the ellipse with respect to the circle is an ellipse given by:
\[ \frac{H^2}{a^2} + \frac{K^2}{\frac{a^4}{b^2}} = 1 \]