The locus of the point of intersection of tangents to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` which meet at right , is

Options

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

Answer

To find the locus of the point of intersection of tangents to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) that meet at right angles, we can follow these steps:

### Step 1: Understand the condition for tangents meeting at right angles
The product of the slopes of two lines that are perpendicular (meet at right angles) is \(-1\). If \(m_1\) and \(m_2\) are the slopes of the tangents, then:
\[ m_1 \cdot m_2 = -1 \]

### Step 2: Write the equation of the tangents to the ellipse
The equation of the tangent to the ellipse in slope-intercept form is given by:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Rearranging this, we can express it as:
\[ y - mx = \pm \sqrt{a^2m^2 + b^2} \]

### Step 3: Square both sides
Squaring both sides gives:
\[ (y - mx)^2 = a^2m^2 + b^2 \]
Expanding the left side:
\[ y^2 - 2ymx + m^2x^2 = a^2m^2 + b^2 \]

### Step 4: Rearranging the equation
Rearranging the equation, we get:
\[ m^2x^2 - 2ymx + (b^2 - y^2) = 0 \]
This is a quadratic equation in \(m\).

### Step 5: Find the product of the roots
For a quadratic equation of the form \(Ax^2 + Bx + C = 0\), the product of the roots is given by \(\frac{C}{A}\). Here, \(A = x^2\), \(B = -2yx\), and \(C = b^2 - y^2\). Thus, the product of the slopes \(m_1\) and \(m_2\) is:
\[ m_1 \cdot m_2 = \frac{b^2 - y^2}{x^2} \]
Setting this equal to \(-1\) (since the tangents meet at right angles):
\[ \frac{b^2 - y^2}{x^2} = -1 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ b^2 - y^2 = -x^2 \]
or
\[ x^2 + y^2 = b^2 \]

### Step 7: Conclusion about the locus
The equation \(x^2 + y^2 = b^2\) represents a circle with center at the origin \((0, 0)\) and radius \(b\).

### Final Answer
The locus of the point of intersection of tangents to the ellipse that meet at right angles is:
\[ x^2 + y^2 = b^2 \]