Any tangent at a point `P (x,y)` to the ellipse `(x^2)/8+(y^2)=1` meets the coordinate axes in the points A and B such that the area of the triangle OAB is least, then the point P is

Options

A. `(sqrt8,0)`
B. `(0,sqrt(18))`
C. `(2,3)`
D. none

Answer

To solve the problem, we need to find the point \( P(x_1, y_1) \) on the ellipse given by the equation

\[ \frac{x^2}{8} + y^2 = 1 \]

such that the area of the triangle formed by the tangent at \( P \) and the coordinate axes is minimized.

### Step 1: Write the equation of the tangent line at point \( P(x_1, y_1) \)

The equation of the tangent to the ellipse at point \( P(x_1, y_1) \) can be expressed as:

\[ \frac{x x_1}{8} + y y_1 = 1 \]

### Step 2: Find the x-intercept and y-intercept of the tangent line

To find the x-intercept \( A \) (where \( y = 0 \)):

\[ \frac{x x_1}{8} + 0 \cdot y_1 = 1 \implies x = \frac{8}{x_1} \]

Thus, the x-intercept \( A \) is \( \left( \frac{8}{x_1}, 0 \right) \).

To find the y-intercept \( B \) (where \( x = 0 \)):

\[ \frac{0 \cdot x_1}{8} + y y_1 = 1 \implies y = \frac{1}{y_1} \]

Thus, the y-intercept \( B \) is \( \left( 0, \frac{1}{y_1} \right) \).

### Step 3: Calculate the area of triangle OAB

The area \( A \) of triangle \( OAB \) can be calculated using the formula:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{8}{x_1} \times \frac{1}{y_1} \]

This simplifies to:

\[ \text{Area} = \frac{4}{x_1 y_1} \]

### Step 4: Substitute \( y_1 \) in terms of \( x_1 \)

From the ellipse equation, we know:

\[ \frac{x_1^2}{8} + y_1^2 = 1 \implies y_1^2 = 1 - \frac{x_1^2}{8} \implies y_1 = \sqrt{1 - \frac{x_1^2}{8}} \]

Substituting \( y_1 \) into the area formula gives:

\[ \text{Area} = \frac{4}{x_1 \sqrt{1 - \frac{x_1^2}{8}}} \]

### Step 5: Minimize the area

To minimize the area, we can differentiate the area with respect to \( x_1 \) and set the derivative to zero:

Let \( A(x_1) = \frac{4}{x_1 \sqrt{1 - \frac{x_1^2}{8}}} \).

Using the quotient rule and simplifying, we find:

\[ \frac{dA}{dx_1} = 0 \]

After differentiating and simplifying, we find critical points. Solving the resulting equation gives:

\[ x_1 = 2 \quad \text{or} \quad x_1 = -2 \]

### Step 6: Find \( y_1 \)

Substituting \( x_1 = 2 \) back into the ellipse equation to find \( y_1 \):

\[ y_1 = \sqrt{1 - \frac{2^2}{8}} = \sqrt{1 - \frac{4}{8}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Conclusion

Thus, the point \( P \) is:

\[ P\left(2, \frac{1}{\sqrt{2}}\right) \]