Answer :
To address the given questions about the curve with the equation [tex]\( x^2 y^2 + x y^4 = 12 \)[/tex]:
### Part (i): Show that [tex]\(\frac{d y}{d x} = -\frac{2 x y + y^3}{2 x^2 + 4 x y^2}\)[/tex]
First, we need to find the derivative [tex]\(\frac{dy}{dx}\)[/tex]. We can use implicit differentiation on the given equation [tex]\( x^2 y^2 + x y^4 = 12 \)[/tex].
1. Differentiate both sides of the equation with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}(x^2 y^2 + x y^4) = \frac{d}{dx}(12) \][/tex]
2. Apply the product rule to differentiate each term on the left-hand side:
[tex]\[ \frac{d}{dx}(x^2 y^2) + \frac{d}{dx}(x y^4) = 0 \][/tex]
For the first term [tex]\( x^2 y^2 \)[/tex]:
[tex]\[ \frac{d}{dx}(x^2 y^2) = 2x y^2 + x^2 \cdot 2y \frac{dy}{dx} \][/tex]
This simplifies to:
[tex]\[ 2x y^2 + 2x^2 y \frac{dy}{dx} \][/tex]
For the second term [tex]\( x y^4 \)[/tex]:
[tex]\[ \frac{d}{dx}(x y^4) = y^4 + x \cdot 4y^3 \frac{dy}{dx} \][/tex]
Now, combining these results:
[tex]\[ 2x y^2 + 2x^2 y \frac{dy}{dx} + y^4 + 4x y^3 \frac{dy}{dx} = 0 \][/tex]
3. Collect terms involving [tex]\(\frac{dy}{dx}\)[/tex] on one side:
[tex]\[ 2x y^2 + y^4 = -2x^2 y \frac{dy}{dx} - 4x y^3 \frac{dy}{dx} \][/tex]
[tex]\[ 2x y^2 + y^4 = \frac{dy}{dx} (-2x^2 y - 4x y^3) \][/tex]
4. Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{2x y^2 + y^4}{-2x^2 y - 4x y^3} \][/tex]
Simplifying the denominator:
[tex]\[ \frac{dy}{dx} = - \frac{2x y^2 + y^3}{2x^2 + 4x y^2} \][/tex]
Therefore, we have shown that:
[tex]\[ \frac{dy}{dx} = -\frac{2 x y + y^3}{2 x^2 + 4 x y^2} \][/tex]
### Part (ii): Find the equation of the tangent to the curve when [tex]\( y = 1 \)[/tex] and [tex]\( x > 0 \)[/tex]
1. Substitute [tex]\( y = 1 \)[/tex] into the original equation to find the value of [tex]\( x \)[/tex]:
[tex]\[ x^2 (1)^2 + x (1)^4 = 12 \][/tex]
[tex]\[ x^2 + x = 12 \][/tex]
Rearrange to solve the quadratic equation:
[tex]\[ x^2 + x - 12 = 0 \][/tex]
2. Solve the quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -12 \)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 48}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm 7}{2} \][/tex]
[tex]\[ x = \frac{6}{2} = 3 \quad \text{(since \( x > 0 \))} \][/tex]
3. Find the slope of the tangent line at [tex]\( (x, y) = (3, 1) \)[/tex]:
Substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = 1 \)[/tex] into the derivative:
[tex]\[ \frac{dy}{dx} = -\frac{2x y + y^3}{2x^2 + 4x y^2} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{(3, 1)} = -\frac{2 \cdot 3 \cdot 1 + 1^3}{2 \cdot 3^2 + 4 \cdot 3 \cdot 1} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{(3, 1)} = -\frac{6 + 1}{18 + 12} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{(3, 1)} = -\frac{7}{30} \][/tex]
4. Write the equation of the tangent line. The point-slope formula for the tangent line is:
[tex]\[ y - y_1 = m (x - x_1) \][/tex]
Where [tex]\((x_1, y_1) = (3, 1)\)[/tex] and [tex]\( m = -\frac{7}{30}\)[/tex]:
[tex]\[ y - 1 = -\frac{7}{30}(x - 3) \][/tex]
Distribute and simplify:
[tex]\[ y - 1 = -\frac{7}{30}x + \frac{7 \cdot 3}{30} \][/tex]
[tex]\[ y - 1 = -\frac{7}{30}x + \frac{21}{30} \][/tex]
[tex]\[ y = -\frac{7}{30}x + \frac{21}{30} + 1 \][/tex]
[tex]\[ y = -\frac{7}{30}x + \frac{21}{30} + \frac{30}{30} \][/tex]
[tex]\[ y = -\frac{7}{30}x + \frac{51}{30} \][/tex]
Therefore, the equation of the tangent line is:
[tex]\[ y = -\frac{7}{30}x + \frac{51}{30} \][/tex]
### Part (i): Show that [tex]\(\frac{d y}{d x} = -\frac{2 x y + y^3}{2 x^2 + 4 x y^2}\)[/tex]
First, we need to find the derivative [tex]\(\frac{dy}{dx}\)[/tex]. We can use implicit differentiation on the given equation [tex]\( x^2 y^2 + x y^4 = 12 \)[/tex].
1. Differentiate both sides of the equation with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}(x^2 y^2 + x y^4) = \frac{d}{dx}(12) \][/tex]
2. Apply the product rule to differentiate each term on the left-hand side:
[tex]\[ \frac{d}{dx}(x^2 y^2) + \frac{d}{dx}(x y^4) = 0 \][/tex]
For the first term [tex]\( x^2 y^2 \)[/tex]:
[tex]\[ \frac{d}{dx}(x^2 y^2) = 2x y^2 + x^2 \cdot 2y \frac{dy}{dx} \][/tex]
This simplifies to:
[tex]\[ 2x y^2 + 2x^2 y \frac{dy}{dx} \][/tex]
For the second term [tex]\( x y^4 \)[/tex]:
[tex]\[ \frac{d}{dx}(x y^4) = y^4 + x \cdot 4y^3 \frac{dy}{dx} \][/tex]
Now, combining these results:
[tex]\[ 2x y^2 + 2x^2 y \frac{dy}{dx} + y^4 + 4x y^3 \frac{dy}{dx} = 0 \][/tex]
3. Collect terms involving [tex]\(\frac{dy}{dx}\)[/tex] on one side:
[tex]\[ 2x y^2 + y^4 = -2x^2 y \frac{dy}{dx} - 4x y^3 \frac{dy}{dx} \][/tex]
[tex]\[ 2x y^2 + y^4 = \frac{dy}{dx} (-2x^2 y - 4x y^3) \][/tex]
4. Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{2x y^2 + y^4}{-2x^2 y - 4x y^3} \][/tex]
Simplifying the denominator:
[tex]\[ \frac{dy}{dx} = - \frac{2x y^2 + y^3}{2x^2 + 4x y^2} \][/tex]
Therefore, we have shown that:
[tex]\[ \frac{dy}{dx} = -\frac{2 x y + y^3}{2 x^2 + 4 x y^2} \][/tex]
### Part (ii): Find the equation of the tangent to the curve when [tex]\( y = 1 \)[/tex] and [tex]\( x > 0 \)[/tex]
1. Substitute [tex]\( y = 1 \)[/tex] into the original equation to find the value of [tex]\( x \)[/tex]:
[tex]\[ x^2 (1)^2 + x (1)^4 = 12 \][/tex]
[tex]\[ x^2 + x = 12 \][/tex]
Rearrange to solve the quadratic equation:
[tex]\[ x^2 + x - 12 = 0 \][/tex]
2. Solve the quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -12 \)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 48}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm 7}{2} \][/tex]
[tex]\[ x = \frac{6}{2} = 3 \quad \text{(since \( x > 0 \))} \][/tex]
3. Find the slope of the tangent line at [tex]\( (x, y) = (3, 1) \)[/tex]:
Substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = 1 \)[/tex] into the derivative:
[tex]\[ \frac{dy}{dx} = -\frac{2x y + y^3}{2x^2 + 4x y^2} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{(3, 1)} = -\frac{2 \cdot 3 \cdot 1 + 1^3}{2 \cdot 3^2 + 4 \cdot 3 \cdot 1} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{(3, 1)} = -\frac{6 + 1}{18 + 12} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{(3, 1)} = -\frac{7}{30} \][/tex]
4. Write the equation of the tangent line. The point-slope formula for the tangent line is:
[tex]\[ y - y_1 = m (x - x_1) \][/tex]
Where [tex]\((x_1, y_1) = (3, 1)\)[/tex] and [tex]\( m = -\frac{7}{30}\)[/tex]:
[tex]\[ y - 1 = -\frac{7}{30}(x - 3) \][/tex]
Distribute and simplify:
[tex]\[ y - 1 = -\frac{7}{30}x + \frac{7 \cdot 3}{30} \][/tex]
[tex]\[ y - 1 = -\frac{7}{30}x + \frac{21}{30} \][/tex]
[tex]\[ y = -\frac{7}{30}x + \frac{21}{30} + 1 \][/tex]
[tex]\[ y = -\frac{7}{30}x + \frac{21}{30} + \frac{30}{30} \][/tex]
[tex]\[ y = -\frac{7}{30}x + \frac{51}{30} \][/tex]
Therefore, the equation of the tangent line is:
[tex]\[ y = -\frac{7}{30}x + \frac{51}{30} \][/tex]
