The locus of the point of intersection of the perpendicular tangents to the ellipse `2x^(2)+3y^(2)=6` is

Options

A. `x^(2)+y^(2)=41`
B. `x^(2)+y^(2)=25`
C. `x^(2)+y^(2)=5`
D. `x^(2)+y^(2)=43`

Answer

To find the locus of the point of intersection of the perpendicular tangents to the ellipse given by the equation \(2x^2 + 3y^2 = 6\), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
We start with the equation of the ellipse:
\[ 2x^2 + 3y^2 = 6 \]
To convert it into standard form, we divide the entire equation by 6:
\[ \frac{x^2}{3} + \frac{y^2}{2} = 1 \]
This shows that the semi-major axis \(a^2 = 3\) and the semi-minor axis \(b^2 = 2\).

### Step 2: Identify the values of \(a^2\) and \(b^2\)
From the standard form, we have:
- \(a^2 = 3\)
- \(b^2 = 2\)

### Step 3: Use the formula for the locus of perpendicular tangents
The locus of the point of intersection of the perpendicular tangents to an ellipse is given by the equation:
\[ x^2 + y^2 = a^2 + b^2 \]
Substituting the values of \(a^2\) and \(b^2\):
\[ x^2 + y^2 = 3 + 2 \]
\[ x^2 + y^2 = 5 \]

### Step 4: Write the final answer
Thus, the locus of the point of intersection of the perpendicular tangents to the ellipse \(2x^2 + 3y^2 = 6\) is:
\[ \boxed{x^2 + y^2 = 5} \]