Maths › Further Pure 1 › Tangents, normals and loci of conics
Tangents, normals and loci of conics
Differentiate a conic through its parameter and the tangent at a general point becomes one reusable equation. Let the point move and the algebra traces out loci.
Builds on Conic sections and Implicit and parametric differentiation.
IN THIS TOPIC
- Find tangents and normals at a general parametric point of any of the four conics.
- Quote the standard tangent forms, and derive them when the question demands it.
- Use and verify the tangency condition for y = mx + c.
- Eliminate the parameter to find the locus of a moving point.
COMMON MISCONCEPTION
A line either obviously crosses a curve or obviously misses it; tangency needs a picture, not algebra.
The tangent at a general point
Parametric differentiation gives the gradient at (at², 2at) on y² = 4ax as 1/t, so the tangent there is ty = x + at² and the normal has gradient −t. One derivation covers every point on the curve. Algebra also detects tangency with no picture at all. Substitute the line into the conic, and a repeated root is a touch while two distinct roots are a crossing.
WORKED EXAMPLE
Tangent by substitution
Show that y = x + 2 is a tangent to y² = 8x, and find the point of contact.
Substitute: (x + 2)² = 8x, so x² − 4x + 4 = 0.
That is (x − 2)² = 0, a repeated root at x = 2, so the line touches instead of crossing.
Contact point (2, 4), which is (at², 2at) with a = 2 and t = 1. The general tangent ty = x + at² reduces to y = x + 2 there, as it should.
The four tangent forms
Each conic has a tangent at its general parametric point, and a question that says 'show that' expects the derivation while a question that says 'hence' expects the quoted result. For the parabola at (at², 2at) it is ty = x + at², with normal y + tx = 2at + at³. For the rectangular hyperbola xy = c² at (ct, c/t) the tangent is x + t²y = 2ct.
The ellipse and hyperbola are tidier than they look. At (a cos t, b sin t) on x²/a² + y²/b² = 1 the tangent is (x/a)cos t + (y/b)sin t = 1. At (a sec t, b tan t) on x²/a² − y²/b² = 1 it is (x/a)sec t − (y/b)tan t = 1. Both come from implicit differentiation in about four lines, so practise the derivation until it is quick.
Tangency conditions for y = mx + c follow from demanding a zero discriminant in the substituted quadratic. You get c = a/m for y² = 4ax, and c² = a²m² + b² for the ellipse, and c² = a²m² − b² for the hyperbola. The parabola condition assumes m is not zero, since a horizontal line meets y² = 4ax exactly once without touching it.
Points that move: loci
A locus question fixes a rule and then lets the point roam. Parametrise the roaming point, express the tracked quantity in terms of the parameter, and eliminate. Whatever survives is the locus's own equation.
WORKED EXAMPLE
Midpoints of parallel chords
Find the locus of midpoints of chords of y² = 8x with gradient 2.
A chord y = 2x + c meets the parabola where y² = 4(y − c), that is y² − 4y + 4c = 0.
The two intersection y-values sum to 4 whatever c is, so the midpoint always has y = 2.
The locus is the horizontal line y = 2, taken inside the parabola. Sliding the chord changes everything except the midpoint's height.
GUIDED PRACTICE
Using the tangency condition
The line y = mx + 3 is a tangent to y² = 12x. Find m and verify by substitution.
Show the working
Here a = 3 and the condition is c = a/m, so 3 = 3/m and m = 1.
Substituting y = x + 3: (x + 3)² = 12x gives x² − 6x + 9 = (x − 3)² = 0.
A double root at x = 3, so the line touches at (3, 6) and confirms the condition it was built from.
ASSESSMENT FOCUS
- Derive gradients parametrically and state dy/dx = (dy/dt)/(dx/dt) before using it.
- A repeated root is the algebraic definition of tangency, so use the word 'repeated' explicitly.
- Quote the general tangent only if you have derived it earlier in the paper, or the question says 'hence'.
- For loci, hunt for combinations of the roots that stay constant; sums and products survive elimination.
- Give the locus as an equation with any restriction stated, since an unrestricted answer often includes points the curve never reaches.
CHECK YOURSELF
Find the gradient of the normal to xy = 9 at the point (3, 3).
Show a hint
Differentiate implicitly or use the parametric form (3t, 3/t).
Show the answer
dy/dx = −9/x², which is −1 at x = 3, so the normal has gradient 1. That normal is the line y = x, an axis of symmetry of the rectangular hyperbola.
Tangent to y² = 4ax at (at², 2at) is ty = x + at²; to xy = c² at (ct, c/t) it is x + t²y = 2ct.
A repeated root in the substituted quadratic means tangency, and a zero discriminant is how the conditions are derived.
For a locus, parametrise the moving point and eliminate the parameter; constants built from the roots survive.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the tangents, normals and loci of conics questions page.
CHECK YOUR PROGRESS
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- Find tangents and normals at a general parametric point of any of the four conics.
- Quote the standard tangent forms, and derive them when the question demands it.
- Use and verify the tangency condition for y = mx + c.
- Eliminate the parameter to find the locus of a moving point.
Open the full revision checklist to see every objective in the course in one place.