Normal Curve 516F53

1. **State the problem:** We have the curve given by $$y = 2x^2 + kx - 3$$ and a point on the curve $(3, -6)$. The normal to the curve at this point is parallel to the line $$x + 5y = 10$$. We need to: - (a) Find the value of $k$. - (b) Find the coordinates where the normal meets the curve again. 2. **Find the slope of the given line:** Rewrite the line equation in slope-intercept form: $$x + 5y = 10 \implies 5y = -x + 10 \implies y = -\frac{1}{5}x + 2$$ So, the slope of the line is $$m = -\frac{1}{5}$$. 3. **Find the slope of the tangent to the curve at $(3, -6)$:** The derivative of the curve is: $$\frac{dy}{dx} = \frac{d}{dx}(2x^2 + kx - 3) = 4x + k$$ At $x=3$: $$m_{tangent} = 4(3) + k = 12 + k$$ 4. **Find the slope of the normal at $(3, -6)$:** The slope of the normal is the negative reciprocal of the tangent slope: $$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{12 + k}$$ 5. **Use the condition that the normal is parallel to the line:** Since the normal is parallel to the line with slope $-\frac{1}{5}$, their slopes are equal: $$-\frac{1}{12 + k} = -\frac{1}{5}$$ Multiply both sides by $-1$: $$\frac{1}{12 + k} = \frac{1}{5}$$ Cross-multiply: $$5 = 12 + k$$ Solve for $k$: $$k = 5 - 12 = -7$$ **Answer for (a):** $$k = -7$$ 6. **Find the coordinates where the normal meets the curve again:** - The curve equation with $k = -7$ is: $$y = 2x^2 - 7x - 3$$ - The slope of the tangent at $x=3$ is: $$m_{tangent} = 12 + (-7) = 5$$ - The slope of the normal is: $$m_{normal} = -\frac{1}{5}$$ - Equation of the normal line at $(3, -6)$: Use point-slope form: $$y - (-6) = -\frac{1}{5}(x - 3)$$ $$y + 6 = -\frac{1}{5}x + \frac{3}{5}$$ $$y = -\frac{1}{5}x + \frac{3}{5} - 6 = -\frac{1}{5}x - \frac{27}{5}$$ 7. **Find the intersection points of the normal line and the curve:** Set the curve equal to the normal line: $$2x^2 - 7x - 3 = -\frac{1}{5}x - \frac{27}{5}$$ Multiply both sides by 5 to clear denominators: $$5(2x^2 - 7x - 3) = 5\left(-\frac{1}{5}x - \frac{27}{5}\right)$$ $$10x^2 - 35x - 15 = -x - 27$$ Bring all terms to one side: $$10x^2 - 35x - 15 + x + 27 = 0$$ $$10x^2 - 34x + 12 = 0$$ 8. **Simplify the quadratic equation:** Divide entire equation by 2: $$5x^2 - 17x + 6 = 0$$ 9. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 5 \cdot 6}}{2 \cdot 5} = \frac{17 \pm \sqrt{289 - 120}}{10} = \frac{17 \pm \sqrt{169}}{10}$$ $$x = \frac{17 \pm 13}{10}$$ Two solutions: - $$x_1 = \frac{17 + 13}{10} = \frac{30}{10} = 3$$ (the original point) - $$x_2 = \frac{17 - 13}{10} = \frac{4}{10} = 0.4$$ 10. **Find the corresponding $y$ for $x=0.4$ on the curve:** $$y = 2(0.4)^2 - 7(0.4) - 3 = 2(0.16) - 2.8 - 3 = 0.32 - 2.8 - 3 = -5.48$$ **Answer for (b):** The normal meets the curve again at $$\boxed{(0.4, -5.48)}$$.