8 The line \( 4 y-3 x=7 \) meets the curve \( 3 x^{2}-8 y=-5 \) at points \( P \) and \( Q \). Find the gradient of the curve at \( P \) and at \( Q \)

To find the gradient of the curve at points \( P \) and \( Q \), we first need to express the line and the curve in a more convenient form and find their points of intersection. The equation of the line is given by: \[ 4y - 3x = 7 \implies y = \frac{3}{4}x + \frac{7}{4} \] The equation of the curve is given by: \[ 3x^2 - 8y = -5 \implies 8y = 3x^2 + 5 \implies y = \frac{3}{8}x^2 + \frac{5}{8} \] Now, we set the two expressions for \( y \) equal to each other to find the points of intersection: \[ \frac{3}{4}x + \frac{7}{4} = \frac{3}{8}x^2 + \frac{5}{8} \] To eliminate the fractions, we can multiply through by 8: \[ 6x + 14 = 3x^2 + 5 \] Rearranging gives us: \[ 3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \] Factoring gives: \[ (x - 3)(x + 1) = 0 \] Thus, \( x = 3 \) or \( x = -1 \). Next, we substitute these \( x \)-values back into the equation of the line to find the corresponding \( y \)-values. For \( x = 3 \): \[ y = \frac{3}{4}(3) + \frac{7}{4} = \frac{9}{4} + \frac{7}{4} = \frac{16}{4} = 4 \] So point \( P = (3, 4) \). For \( x = -1 \): \[ y = \frac{3}{4}(-1) + \frac{7}{4} = -\frac{3}{4} + \frac{7}{4} = \frac{4}{4} = 1 \] So point \( Q = (-1, 1) \). Next, we find the gradient of the curve at these points by taking the derivative of the curve's equation \( y \): The derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{3}{8}x^2 + \frac{5}{8}\right) = \frac{3}{4}x \] Now we can find the gradients at points \( P \) and \( Q \). At point \( P \) where \( x = 3 \): \[ \frac{dy}{dx} = \frac{3}{4}(3) = \frac{9}{4} \] At point \( Q \) where \( x = -1 \): \[ \frac{dy}{dx} = \frac{3}{4}(-1) = -\frac{3}{4} \] Thus, the gradients of the curve at points \( P \) and \( Q \) are \( \frac{9}{4} \) and \( -\frac{3}{4} \), respectively.