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 a conic.
- Use and verify the tangency condition for y = mx + c against each conic.
- Eliminate the parameter to find the locus of a moving point.
WHAT YOU PROBABLY THINK
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, every point. Algebra also detects tangency without any picture, sinking the opening claim: substitute the line into the conic, and a repeated root is a touch while 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 rather than crosses.
Contact point (2, 4), which is (at², 2at) with a = 2, t = 1, and the general tangent ty = x + at² reduces to y = x + 2 there, as it should.
The tangency condition for y = mx + c is worth knowing outright: c = a/m for y² = 4ax, and c² = a²m² + b² for the ellipse. Both come from demanding a zero discriminant in the substituted quadratic.
Points that move: loci
A locus question fixes a rule and lets the point roam. Parametrise the roaming point, express the tracked quantity in the parameter, then eliminate the parameter: what remains 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 (inside the parabola): sliding the chord changes everything except the midpoint's height.
YOUR TURN
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: 3 = 3/m, so m = 1.
Substituting y = x + 3: (x + 3)² = 12x gives x² − 6x + 9 = (x − 3)² = 0.
A double root at x = 3: the line touches at (3, 6), confirming the condition it was built from.
THE EXAM BIT
- Derive gradients parametrically: dy/dx = (dy/dt)/(dx/dt), stated before use.
- A repeated root is the algebraic definition of tangency; say the word 'repeated' explicitly.
- For loci, hunt for combinations of roots that stay constant: sums and products survive elimination.
- Quote the general tangent ty = x + at² only after deriving it once in the paper.
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
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Tangent to y² = 4ax at (at², 2at): ty = x + at²; a repeated root in the substituted quadratic means tangency.
For a locus, parametrise the moving point and eliminate the parameter; constants of 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.
CHECK YOUR PROGRESS
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- Find tangents and normals at a general parametric point of a conic.
- Use and verify the tangency condition for y = mx + c against each conic.
- 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.
No animated video for this topic yet; these notes stand alone.