Practice Questions: Tangents & Normals to Curves

$$\begin{align*} \text{Step 1: Find the point} \\ f(2) &= (2)^2 = 4 \\ \text{Point: } &(2; 4) \\ \\ \text{Step 2: Find } f'(x) \\ f'(x) &= 2x \\ \\ \text{Step 3: Find gradient at } x = 2 \\ m &= f'(2) = 2(2) = 4 \\ \\ \text{Step 4: Use } y - y_1 &= m(x - x_1) \\ y - 4 &= 4(x - 2) \\ y - 4 &= 4x - 8 \\ y &= 4x - 4 \end{align*}$$

$$\begin{align*} \frac{dy}{dx} &= 6x - 2 \\ \\ \text{At } x = 1: \\ m &= 6(1) - 2 \\ &= 4 \end{align*}$$

Answer: The gradient is 4.

$$\begin{align*} m_{\text{tangent}} \times m_{\text{normal}} &= -1 \\ -2 \times m_{\text{normal}} &= -1 \\ m_{\text{normal}} &= \frac{-1}{-2} \\ &= \frac{1}{2} \end{align*}$$

Answer: The gradient of the normal is $\frac{1}{2}$.

$$\begin{align*} \text{Step 1: Find the point} \\ f(2) &= (2)^3 + 2(2)^2 - 7(2) + 1 \\ &= 8 + 8 - 14 + 1 \\ &= 3 \\ \text{Point: } &(2; 3) \\ \\ \text{Step 2: Find } f'(x) \\ f'(x) &= 3x^2 + 4x - 7 \\ \\ \text{Step 3: Find gradient} \\ m &= f'(2) = 3(2)^2 + 4(2) - 7 \\ &= 12 + 8 - 7 \\ &= 13 \\ \\ \text{Step 4: Equation} \\ y - 3 &= 13(x - 2) \\ y &= 13x - 26 + 3 \\ y &= 13x - 23 \end{align*}$$

$$\begin{align*} \text{Parallel lines have equal gradients} \\ \text{The line } y = 4x - 2 \text{ has gradient } m = 4 \\ \\ \text{Expand } f(x): \\ f(x) &= (2x - 1)^2 \\ &= 4x^2 - 4x + 1 \\ \\ f'(x) &= 8x - 4 \\ \\ \text{Set } f'(x) = 4: \\ 8x - 4 &= 4 \\ 8x &= 8 \\ x &= 1 \\ \\ \text{Find y-coordinate:} \\ f(1) &= (2(1) - 1)^2 = 1 \\ \\ \text{Point: } &(1; 1) \end{align*}$$

$$\begin{align*} \text{Y-intercept is where } x = 0: \\ y &= -0^2 + 4(0) - 3 = -3 \\ \text{Point: } &(0; -3) \\ \\ \frac{dy}{dx} &= -2x + 4 \\ \\ \text{Gradient of tangent at } x = 0: \\ m_{\text{tangent}} &= -2(0) + 4 = 4 \\ \\ \text{Gradient of normal:} \\ m_{\text{normal}} &= -\frac{1}{4} \\ \\ \text{Equation of normal:} \\ y - (-3) &= -\frac{1}{4}(x - 0) \\ y + 3 &= -\frac{1}{4}x \\ y &= -\frac{1}{4}x - 3 \end{align*}$$

$$\begin{align*} \text{Rewrite and differentiate:} \\ f(x) &= x^{-1} \\ f'(x) &= -x^{-2} = -\frac{1}{x^2} \\ \\ \text{Gradient at } x = 2: \\ m &= -\frac{1}{(2)^2} = -\frac{1}{4} \\ \\ \text{Equation of tangent:} \\ y - \frac{1}{2} &= -\frac{1}{4}(x - 2) \\ y - \frac{1}{2} &= -\frac{1}{4}x + \frac{1}{2} \\ y &= -\frac{1}{4}x + 1 \\ \\ \text{X-intercept (set } y = 0\text{):} \\ 0 &= -\frac{1}{4}x + 1 \\ \frac{1}{4}x &= 1 \\ x &= 4 \end{align*}$$

Answer: Tangent is $y = -\frac{1}{4}x + 1$, x-intercept at $(4; 0)$.

$$\begin{align*} \text{Gradient of given line:} \\ 2y + x - 4 &= 0 \\ 2y &= -x + 4 \\ y &= -\frac{1}{2}x + 2 \\ \text{Gradient} &= -\frac{1}{2} \\ \\ \text{For perpendicular line:} \\ m_{\text{tangent}} &= -\frac{1}{-1/2} = 2 \\ \\ \text{Expand and differentiate:} \\ f(x) &= 4x^2 - 4x + 1 \\ f'(x) &= 8x - 4 \\ \\ \text{Find where } f'(x) = 2: \\ 8x - 4 &= 2 \\ 8x &= 6 \\ x &= \frac{3}{4} \\ \\ \text{Find y-coordinate:} \\ f\left(\frac{3}{4}\right) &= \left(2\left(\frac{3}{4}\right) - 1\right)^2 \\ &= \left(\frac{1}{2}\right)^2 = \frac{1}{4} \\ \\ \text{Point: } &\left(\frac{3}{4}; \frac{1}{4}\right) \end{align*}$$

$$\begin{align*} \text{Find the point:} \\ y &= (1)^3 - 3(1)^2 + 2 \\ &= 1 - 3 + 2 = 0 \\ \text{Point: } &(1; 0) \\ \\ \text{Find derivative:} \\ \frac{dy}{dx} &= 3x^2 - 6x \\ \\ \text{Gradient at } x = 1: \\ m &= 3(1)^2 - 6(1) = -3 \\ \\ \text{Tangent equation:} \\ y - 0 &= -3(x - 1) \\ y &= -3x + 3 \\ \\ \text{Normal gradient:} \\ m_{\text{normal}} &= -\frac{1}{-3} = \frac{1}{3} \\ \\ \text{Normal equation:} \\ y - 0 &= \frac{1}{3}(x - 1) \\ y &= \frac{1}{3}x - \frac{1}{3} \end{align*}$$

Answer: Tangent: $y = -3x + 3$, Normal: $y = \frac{1}{3}x - \frac{1}{3}$

$$\begin{align*} \text{Since the curve passes through } (1; 3): \\ 3 &= a(1)^2 + b(1) \\ 3 &= a + b \quad \text{...(1)} \\ \\ \text{Find derivative:} \\ \frac{dy}{dx} &= 2ax + b \\ \\ \text{At } x = 1, \text{ gradient is 8:} \\ 8 &= 2a(1) + b \\ 8 &= 2a + b \quad \text{...(2)} \\ \\ \text{Subtract (1) from (2):} \\ 8 - 3 &= (2a + b) - (a + b) \\ 5 &= a \\ \\ \text{Substitute into (1):} \\ 3 &= 5 + b \\ b &= -2 \end{align*}$$

Answer: $a = 5$ and $b = -2$

$$\begin{align*} \text{Step 1: Find y-coordinate} \\ g(-1) &= (-1 + 2)[2(-1) + 1]^2 \\ &= (1)(-1)^2 \\ &= 1 \\ \text{Point: } &(-1; 1) \\ \\ \text{Step 2: Expand the function} \\ g(x) &= (x + 2)(2x + 1)^2 \\ &= (x + 2)(4x^2 + 4x + 1) \\ &= 4x^3 + 4x^2 + x + 8x^2 + 8x + 2 \\ &= 4x^3 + 12x^2 + 9x + 2 \\ \\ \text{Step 3: Find derivative} \\ g'(x) &= 12x^2 + 24x + 9 \\ \\ \text{Step 4: Find gradient} \\ g'(-1) &= 12(-1)^2 + 24(-1) + 9 \\ &= 12 - 24 + 9 \\ &= -3 \\ \\ \text{Step 5: Equation} \\ y - 1 &= -3(x - (-1)) \\ y - 1 &= -3(x + 1) \\ y - 1 &= -3x - 3 \\ y &= -3x - 2 \end{align*}$$

$$\begin{align*} \text{Step 1: Make y the subject} \\ xy &= -4 \\ y &= -\frac{4}{x} = -4x^{-1} \\ \\ \text{Step 2: Differentiate} \\ \frac{dy}{dx} &= -4(-1x^{-2}) \\ &= 4x^{-2} = \frac{4}{x^2} \\ \\ \text{Step 3: Gradient of tangent at } x = -1: \\ \frac{dy}{dx} &= \frac{4}{(-1)^2} = 4 \\ \\ \text{Step 4: Gradient of normal} \\ m_{\text{normal}} &= -\frac{1}{4} \\ \\ \text{Step 5: Equation of normal} \\ y - 4 &= -\frac{1}{4}(x - (-1)) \\ y - 4 &= -\frac{1}{4}(x + 1) \\ y - 4 &= -\frac{1}{4}x - \frac{1}{4} \\ y &= -\frac{1}{4}x + \frac{15}{4} \end{align*}$$