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Differentiation help (1 Viewer)
- Thread starter csi
- Start date
Your problem is to find the point, P say, on the parabola y = f(x) where the normal passes. This normal has gradient -(1/4), so that the gradient of the tangent to the parabola at P is 4. You use this fact to find P, using f'(x) = 4.
 = 4x-5 = 4 \implies x = \frac {9}{4} \implies y = f(\frac{9}{4}) = \frac {7}{8} \\ \\ \therefore $ co-ords of P is $ (\frac {9}{4},\frac {7}{8})\\ \\ \therefore $eqn of normal is: $ y - \frac {7}{8} = -\frac{1}{4}(x - \frac {9}{4} )\\ \\ $ x 16: $ 16y - 14 = -4x + 9 \implies 4x + 16y - 23 = 0)
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makes sense, thank you so muchYour problem is to find the point, P say, on the parabola y = f(x) where the normal passes. This normal has gradient -(1/4), so that the gradient of the tangent to the parabola at P is 4. You use this to find P, using f'(x) = 4.
