Finding Where a Tangent Is Parallel to a Line

“Parallel” is the keyword: parallel lines have equal gradients, so the tangent’s gradient equals the given line’s gradient. Set $\frac{dy}{dx}$ equal to that gradient and solve. It links differentiation to straight-line gradients.

Find the coordinates of the point on the curve $y = x^2 - 4x + 7$ at which the tangent is parallel to the line $y = 2x - 3$. [5]

The working

Step 1, read the line’s gradient. $y = 2x - 3$ has gradient $2$, so the tangent’s gradient must also be $2$.

Step 2, differentiate the curve: $\frac{dy}{dx} = 2x - 4 \quad \text{(M1)}$

Step 3, set $\frac{dy}{dx}$ equal to the required gradient and solve: $2x - 4 = 2 \;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3 \quad \text{(M1, A1)}$

Step 4, find $y$ from the original curve at $x = 3$: $y = 3^2 - 4(3) + 7 = 9 - 12 + 7 = 4 \quad \text{(M1, A1)}$

The point is $(3, 4)$.

Where the marks are won and lost

• Parallel means equal gradient. Setting $\frac{dy}{dx} = 2$ is the key step. (If it said perpendicular, you’d set it to the negative reciprocal $-\frac12$ instead.)
• The $y$-coordinate comes from the curve, not the derivative. Substituting $x = 3$ into $\frac{dy}{dx}$ gives the gradient again, not $y$.
• Give both coordinates, the question asks for the point.

Common mistakes

• Setting $\frac{dy}{dx}$ to the line’s $y$-intercept ($-3$) instead of its gradient.
• Finding $x$ but forgetting the $y$-coordinate.
• Confusing parallel (equal gradient) with perpendicular (negative reciprocal).

Full method: Tangents & Normals notes. Topic home: Calculus pillar.