Does the straight line `(x)/(a)+(y)/(b)=2` touch the ellipse `(x/a)^(2)+(y/b)^(2)=2?` justify.

Answer

To determine if the straight line \(\frac{x}{a} + \frac{y}{b} = 2\) touches the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 2\), we will follow a systematic approach.

### Step 1: Rewrite the equations

The given line can be rewritten in the slope-intercept form:

\[ \frac{y}{b} = 2 - \frac{x}{a} \]
\[ y = -\frac{b}{a}x + 2b \]

The given ellipse can be rewritten as:

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

### Step 2: Identify the slope of the line

From the line equation \(y = -\frac{b}{a}x + 2b\), we can identify the slope \(m\) of the line:

\[ m = -\frac{b}{a} \]

### Step 3: Use the condition for tangency

For the line to touch the ellipse, the distance \(c\) from the center of the ellipse to the line must equal the semi-minor axis of the ellipse. The formula for the distance \(c\) from the origin (0, 0) to the line \(Ax + By + C = 0\) is:

\[ c = \frac{|C|}{\sqrt{A^2 + B^2}} \]

Rearranging the line equation \(\frac{x}{a} + \frac{y}{b} = 2\) gives:

\[ \frac{x}{a} + \frac{y}{b} - 2 = 0 \]

Here, \(A = \frac{1}{a}\), \(B = \frac{1}{b}\), and \(C = -2\).

### Step 4: Calculate the distance \(c\)

Substituting the values into the distance formula:

\[ c = \frac{|-2|}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2}} = \frac{2}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \]

### Step 5: Set the distance equal to the semi-minor axis

For the ellipse, the semi-major and semi-minor axes are given by:

- Semi-major axis: \(\sqrt{2}a\)
- Semi-minor axis: \(\sqrt{2}b\)

We need to check if:

\[ c^2 = 4b^2 \]

Calculating \(c^2\):

\[ c^2 = \left(\frac{2}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}\right)^2 = \frac{4}{\frac{1}{a^2} + \frac{1}{b^2}} = 4 \cdot \frac{a^2b^2}{b^2 + a^2} \]

### Step 6: Compare with \(4b^2\)

For tangency, we need:

\[ \frac{4a^2b^2}{b^2 + a^2} = 4b^2 \]

Dividing both sides by 4:

\[ \frac{a^2b^2}{b^2 + a^2} = b^2 \]

Cross multiplying gives:

\[ a^2b^2 = b^2(b^2 + a^2) \]

This simplifies to:

\[ a^2b^2 = b^4 + a^2b^2 \]

Subtracting \(a^2b^2\) from both sides results in:

\[ 0 = b^4 \]

This implies that \(b = 0\), which is not a valid case for an ellipse. Thus, the line does indeed touch the ellipse.

### Conclusion

The line \(\frac{x}{a} + \frac{y}{b} = 2\) touches the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 2\) at the point \((a, b)\).