d) Find the equation of the tangent and normal line to the curve 12(x^2+y^2)=25xy at the point (3,4) (3 Marks) e) A body moves in a straight line so that its distance S meters from a

To find the equation of the tangent and normal lines to the curve \(12(x^2 + y^2) = 25xy\) at the point \((3, 4)\), we need to follow these steps:<br /><br />1. **Implicit Differentiation**: Differentiate the given equation with respect to \(x\).<br />2. **Evaluate the Derivative at the Given Point**: Find the slope of the tangent line at \((3, 4)\).<br />3. **Equation of the Tangent Line**: Use the point-slope form of the equation of a line.<br />4. **Equation of the Normal Line**: The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent line's slope.<br /><br />Let's go through these steps in detail:<br /><br />### Step 1: Implicit Differentiation<br /><br />Given:<br />\[ 12(x^2 + y^2) = 25xy \]<br /><br />Differentiate both sides with respect to \(x\):<br />\[ 12 \cdot 2x + 12 \cdot 2y \frac{dy}{dx} = 25y + 25x \frac{dy}{dx} \]<br />\[ 24x + 24y \frac{dy}{dx} = 25y + 25x \frac{dy}{dx} \]<br /><br />Rearrange to solve for \(\frac{dy}{dx}\):<br />\[ 24y \frac{dy}{dx} - 25x \frac{dy}{dx} = 25y - 24x \]<br />\[ \frac{dy}{dx} (24y - 25x) = 25y - 24x \]<br />\[ \frac{dy}{dx} = \frac{25y - 24x}{24y - 25x} \]<br /><br />### Step 2: Evaluate the Derivative at the Given Point<br /><br />Substitute \(x = 3\) and \(y = 4\) into the derivative:<br />\[ \frac{dy}{dx} \bigg|_{(3, 4)} = \frac{25 \cdot 4 - 24 \cdot 3}{24 \cdot 4 - 25 \cdot 3} \]<br />\[ = \frac{100 - 72}{96 - 75} \]<br />\[ = \frac{28}{21} \]<br />\[ = \frac{4}{3} \]<br /><br />So, the slope of the tangent line at \((3, 4)\) is \(\frac{4}{3}\).<br /><br />### Step 3: Equation of the Tangent Line<br /><br />Using the point-slope form of the equation of a line:<br />\[ y - y_1 = m(x - x_1) \]<br />where \(m\) is the slope and \((x_1, y_1)\) is the point \((3, 4)\):<br />\[ y - 4 = \frac{4}{3}(x - 3) \]<br /><br />Simplify:<br />\[ y - 4 = \frac{4}{3}x - 4 \]<br />\[ y = \frac{4}{3}x - 4 + 4 \]<br />\[ y = \frac{4}{3}x \]<br /><br />### Step 4: Equation of the Normal Line<br /><br />The slope of the normal line is the negative reciprocal of the slope of the tangent line:<br />\[ m_{\text{normal}} = -\frac{3}{4} \]<br /><br />Using the point-slope form again:<br />\[ y - 4 = -\frac{3}{4}(x - 3) \]<br /><br />Simplify:<br />\[ y - 4 = -\frac{3}{4}x + \frac{9}{4} \]<br />\[ y = -\frac{3}{4}x + \frac{9}{4} + 4 \]<br />\[ y = -\frac{3}{4}x + \frac{9}{4} + \frac{16}{4} \]<br />\[ y = -\frac{3}{4}x + \frac{25}{4} \]<br /><br />So, the equations are:<br />- Tangent line: \( y = \frac{4}{3}x \)<br />- Normal line: \( y = -\frac{3}{4}x + \frac{25}{4} \)<br /><br />### Summary<br /><br />The equation of the tangent line to the curve \(12(x^2 + y^2) = 25xy\) at the point \((3, 4)\) is:<br />\[ y = \frac{4}{3}x \]<br /><br />The equation of the normal line to the curve at the same point is:<br />\[ y = -\frac{3}{4}x + \frac{25}{4} \]