Find the equation of the tangent to the curve `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` at the point `(sqrt(2)a, b)`.

Answer

To find the equation of the tangent to the curve \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) at the point \((\sqrt{2}a, b)\), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation of the curve:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) - \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = 0 \]
Using the chain rule, we have:
\[ \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ \frac{2y}{b^2} \frac{dy}{dx} = \frac{2x}{a^2} \]
Thus,
\[ \frac{dy}{dx} = \frac{x}{a^2} \cdot \frac{b^2}{y} \]

### Step 3: Substitute the point into the derivative
We need to evaluate \(\frac{dy}{dx}\) at the point \((\sqrt{2}a, b)\):
\[ \frac{dy}{dx} \bigg|_{(\sqrt{2}a, b)} = \frac{\sqrt{2}a}{a^2} \cdot \frac{b^2}{b} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\sqrt{2}}{a} \cdot b = \frac{\sqrt{2}b}{a} \]

### Step 4: Use the point-slope form of the tangent line
The equation of the tangent line in point-slope form is:
\[ y - y_0 = m(x - x_0) \]
where \(m\) is the slope \(\frac{dy}{dx}\) and \((x_0, y_0)\) is the point \((\sqrt{2}a, b)\). Substituting the values:
\[ y - b = \frac{\sqrt{2}b}{a}(x - \sqrt{2}a) \]

### Step 5: Rearranging the equation
Expanding and rearranging gives:
\[ y - b = \frac{\sqrt{2}b}{a}x - \frac{\sqrt{2}b}{a}\sqrt{2}a \]
\[ y - b = \frac{\sqrt{2}b}{a}x - 2b \]
\[ y = \frac{\sqrt{2}b}{a}x - 2b + b \]
\[ y = \frac{\sqrt{2}b}{a}x - b \]

### Step 6: Final form of the equation
Rearranging to get the standard form:
\[ \frac{\sqrt{2}b}{a}x - y - b = 0 \]

Thus, the equation of the tangent to the curve at the point \((\sqrt{2}a, b)\) is:
\[ \sqrt{2}bx - ay - ab = 0 \]