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Tangents, normals and loci of conics questions
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.
6 original questions · 23 marks · the tangents, normals and loci of conics notes · Further Pure 1
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Write down the equation of the tangent to y² = 4ax at the point (at², 2at), and state the gradient of the normal there.
Worked answer
The tangent is ty = x + at², with gradient 1/t. The normal is perpendicular, so its gradient is −t. B1 for the tangent, B1 for the normal gradient.Find the equation of the tangent to y² = 16x at the point where t = 2.
Worked answer
Here 4a = 16 so a = 4, and the point is (at², 2at) = (16, 16). The tangent ty = x + at² becomes 2y = x + 16, that is y = x/2 + 8. Checking: 16² = 256 = 16 × 16, so the point really is on the curve. B1 for a = 4, M1 for substituting into ty = x + at², A1 for 2y = x + 16.Show that y = 2x + 2 is a tangent to y² = 16x, and find the point of contact.
Worked answer
Substituting: (2x + 2)² = 16x gives 4x² − 8x + 4 = 0, that is 4(x − 1)² = 0. The repeated root x = 1 means the line touches rather than crosses, at the point (1, 4). M1 for substituting, A1 for 4x² − 8x + 4 = 0, A1 for the repeated root, A1 for (1, 4). The condition c = a/m agrees: a = 4 and m = 2 give c = 2.Find the equations of the tangent and the normal to xy = 16 at the point (4, 4).
Worked answer
Differentiating implicitly: y + x dy/dx = 0, so dy/dx = −y/x = −1 at (4, 4). The tangent is y − 4 = −(x − 4), that is y = 8 − x. The normal has gradient 1: y = x, which is the hyperbola's own axis of symmetry through that point. M1 for differentiating implicitly, A1 for dy/dx = −1, A1 for y = 8 − x, A1 for y = x.Find the locus of the midpoints of chords of y² = 16x that have gradient 4.
Worked answer
A chord y = 4x + c meets the curve where y² = 4(y − c), that is y² − 4y + 4c = 0. The two y-values always sum to 4 regardless of c, so the midpoint has y = 2 every time. The locus is the line y = 2, taken inside the parabola. M1 for substituting y = 4x + c, A1 for y² − 4y + 4c = 0, M1 for the sum of the roots, A1 for y = 2.The point P(at², 2at) lies on the parabola y² = 4ax, where t ≠ 0. The normal at P meets the parabola again at Q(aT², 2aT). Show that T = −t − 2/t. Hence find Q when a = 4 and t = 1.
Worked answer
The tangent at P is ty = x + at², of gradient 1/t, so the normal has gradient −t.
Its equation is y − 2at = −t(x − at²), that is y + tx = 2at + at³.
Q lies on it, so 2aT + atT² = 2at + at³. Divide by a, which is never zero:
2T + tT² = 2t + t³.
Collect everything on one side and factorise, expecting T = t to be a root, since P itself is on both the normal and the curve:
t(T² − t²) + 2(T − t) = 0, so (T − t)[t(T + t) + 2] = 0.
Q is not P, so T ≠ t and the second bracket must vanish: t(T + t) = −2, giving T + t = −2/t and T = −t − 2/t as required. Spotting the factor (T − t) is what turns a cubic into one line; expanding and solving from scratch is where this question is lost.
With a = 4 and t = 1: T = −1 − 2 = −3, so Q = (aT², 2aT) = (36, −24). Check: (−24)² = 576 = 16 × 36, and the normal y + x = 12 does pass through (36, −24). M1 for the normal gradient, A1 for its equation, M1 for substituting Q, M1 for factorising out (T − t), A1 for the printed result, A1 for (36, −24).
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