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: (a) parabola (b) ellipse (c) hyperbola (d) circle

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 can follow these steps:

### Step 1: Understand the equations of the conics
The ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The circle is given by:
\[ x^2 + y^2 = a^2 \]

### Step 2: Write the equation of the tangent to the ellipse
The equation of the tangent to the ellipse at point \((x_0, y_0)\) can be expressed in slope form as:
\[ y = mx + \sqrt{a^2 m^2 + b^2} \]
where \(m\) is the slope of the tangent line.

### Step 3: Set up the equation for the pole
Let the pole of the tangent line with respect to the circle be \(P(h, k)\). The equation of the circle can be rewritten in terms of the pole:
\[ x h + y k = a^2 \]

### Step 4: Substitute the tangent equation into the pole equation
Substituting the tangent equation into the pole equation gives:
\[ x h + (mx + \sqrt{a^2 m^2 + b^2}) k = a^2 \]
This simplifies to:
\[ hx + kmx + k\sqrt{a^2 m^2 + b^2} = a^2 \]

### Step 5: Rearranging the equation
Rearranging the above equation, we can isolate terms involving \(m\):
\[ (h + km)x + k\sqrt{a^2 m^2 + b^2} = a^2 \]

### Step 6: Finding the relationship between \(h\) and \(k\)
From the equation \(k\sqrt{a^2 m^2 + b^2} = a^2 - (h + km)x\), we can square both sides and simplify to find a relationship between \(h\) and \(k\):
\[ \frac{h^2}{a^2} + \frac{k^2}{b^2} = 1 \]
This is the equation of an ellipse.

### Final Result
Thus, the locus of the poles of the tangents to the ellipse with respect to the circle is:
\[ \frac{h^2}{a^2} + \frac{k^2}{b^2} = 1 \]
This indicates that the locus is an ellipse.

### Answer
The answer is (b) ellipse.