Questions on tangents, normals and loci of conics

Question 1

Write down the equation of the tangent to y² = 4ax at the point (at², 2at), and state the gradient of the normal there.

Accepted 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.

Question 2

Find the equation of the tangent to y² = 16x at the point where t = 2.

Accepted 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.

Question 3

Write down the equation of the tangent to the rectangular hyperbola xy = c² at the point (ct, c/t), and the equation of the tangent to the ellipse x²/a² + y²/b² = 1 at the point (a cos t, b sin t).

Accepted Answer

Rectangular hyperbola: x + t²y = 2ct. Ellipse: (x/a) cos t + (y/b) sin t = 1. B1 B1. Both are worth carrying in memory, and both can be rebuilt in three lines from dy/dx if memory fails: the hyperbola's gradient at the point is −1/t², the ellipse's is −(b cos t)/(a sin t).

Question 4

State the condition on c for the line y = mx + c to be a tangent to the parabola y² = 4ax, and the condition for it to be a tangent to the ellipse x²/a² + y²/b² = 1.

Accepted Answer

Parabola: c = a/m. Ellipse: c² = a²m² + b². B1 B1. Each comes from substituting the line into the curve and demanding a repeated root; the hyperbola's version, c² = a²m² − b², differs from the ellipse's only in the sign that separates the two curves' equations.

Question 5

Show that y = 2x + 2 is a tangent to y² = 16x, and find the point of contact.

Accepted 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.

Question 6

Find the equations of the tangent and the normal to xy = 16 at the point (4, 4).

Accepted 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.

Question 7

Find the locus of the midpoints of chords of y² = 16x that have gradient 4.

Accepted 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.

Question 8

The point P(3 cos t, 2 sin t) lies on the ellipse x²/9 + y²/4 = 1. Show that the tangent at P is (x/3) cos t + (y/2) sin t = 1, and find the equation of the tangent at the point where t = π/3, giving integer coefficients.

Accepted Answer

Differentiate parametrically: dx/dt = −3 sin t and dy/dt = 2 cos t, so dy/dx = −(2 cos t)/(3 sin t). The tangent is y − 2 sin t = −(2 cos t)/(3 sin t) (x − 3 cos t); multiplying through by 3 sin t and expanding gives 2x cos t + 3y sin t = 6 cos²t + 6 sin²t = 6, and dividing by 6 gives (x/3) cos t + (y/2) sin t = 1, with sin²t + cos²t = 1 doing the tidying.
At t = π/3, cos t = 1/2 and sin t = √3/2, so x/6 + √3y/4 = 1, and multiplying by 12 gives 2x + 3√3y = 12. The point of contact (3/2, √3) checks: 3 + 9 = 12. Substituting the line back into the ellipse collapses to 4(2x − 3)² = 0, a repeated root, which is tangency in algebraic form. M1 for the parametric gradient, A1 for dy/dx, M1 for forming the line and using sin²t + cos²t = 1, A1 for the printed tangent, A1 for 2x + 3√3y = 12.

Question 9

The line y = 2x + c is a tangent to the hyperbola x²/4 − y²/9 = 1. Find the possible values of c, and verify your answer against the condition c² = a²m² − b².

Accepted Answer

Substitute the line into the hyperbola and clear fractions: 9x² − 4(2x + c)² = 36, which collects to 7x² + 16cx + 4c² + 36 = 0, taking care with the sign when the bracket is expanded and subtracted.
Tangency needs a repeated root, so the discriminant vanishes: (16c)² − 4 × 7 × (4c² + 36) = 144c² − 1008 = 0, giving c² = 7 and c = ±√7.
The condition agrees: a² = 4, b² = 9 and m = 2 give c² = 4 × 4 − 9 = 7. A line of gradient m can only touch when a²m² > b², that is |m| > 3/2, the asymptote gradient, and gradient 2 clears it. M1 for substituting the line, A1 for 7x² + 16cx + 4c² + 36 = 0, M1 for setting the discriminant to zero, A1 for c = ±√7, B1 for the check against c² = a²m² − b².

Question 10

The point P(at², 2at) lies on the parabola y² = 4ax. By differentiating parametrically, derive the equation of the tangent at P in the form ty = x + at².

Accepted Answer

dx/dt = 2at and dy/dt = 2a, so dy/dx = (dy/dt)/(dx/dt) = 2a/(2at) = 1/t. M1 for the parametric derivatives, A1 for the gradient 1/t.
The tangent is y − 2at = (1/t)(x − at²). M1 for the line through P with that gradient. Multiplying by t: ty − 2at² = x − at², so ty = x + at². A1. Implicit differentiation of y² = 4ax gives 2y dy/dx = 4a and the same gradient 2a/y = 1/t; either route is acceptable when the question asks for a derivation, but quoting the result is not.

Question 11

Show that the normal to the rectangular hyperbola xy = c² at the point P(ct, c/t), where t ≠ 0, has equation t³x − ty = c(t⁴ − 1).

Accepted Answer

Differentiating xy = c² implicitly, y + x dy/dx = 0, so dy/dx = −y/x = −(c/t)/(ct) = −1/t². M1 for the gradient of the tangent. The normal is perpendicular, with gradient t². A1.
Its equation is y − c/t = t²(x − ct). M1. Multiplying by t: ty − c = t³x − ct⁴, so t³x − ty = c(t⁴ − 1). A1. Check at t = 1: the normal at (c, c) is x − y = 0, the line y = x through the origin, which is the hyperbola's axis of symmetry there. Writing the normal gradient as 1/t² instead of t² negates the tangent gradient without inverting it, and the printed form never appears.

Question 12

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.

Accepted 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.
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).

Question 13

The tangent to the rectangular hyperbola xy = 9 at the point P(3t, 3/t), where t ≠ 0, meets the x-axis at A and the y-axis at B. Show that the area of triangle OAB, where O is the origin, is independent of t, and find the locus of the midpoint of AB.

Accepted Answer

Differentiating implicitly, y + x dy/dx = 0, so at P the gradient is −y/x = −(3/t)/(3t) = −1/t². The tangent is y − 3/t = −(1/t²)(x − 3t), which tidies to x + t²y = 6t, the standard x + t²y = 2ct with c = 3.
Setting y = 0 gives A(6t, 0); setting x = 0 gives B(0, 6/t). The triangle OAB is right-angled at O, so its area is ½ |6t| |6/t| = 18 for every t.
The midpoint of AB is (3t, 3/t), which is P itself: the point of contact bisects the intercepted segment. Writing x = 3t and y = 3/t and eliminating t gives xy = 9, so the locus of the midpoint is the hyperbola xy = 9 itself, every point of it being reached as t runs over the nonzero reals. M1 for the gradient at P, A1 for x + t²y = 6t, A1 for both intercepts, A1 for the constant area 18, M1 for the midpoint and eliminating t, A1 for the locus xy = 9.

Question 14

The parabola y² = 4ax has focus S(a, 0), and P(at², 2at) is a point on it. Show that the foot of the perpendicular from S to the tangent at P lies on the y-axis, and find its coordinates. Find also the locus of the midpoint of SP as t varies.

Accepted 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.
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).

Question 15

P(cp, c/p) and Q(cq, c/q) are two points on the rectangular hyperbola xy = c², with p ≠ q and pq ≠ 0. Show that the chord PQ has equation x + pqy = c(p + q), and deduce the equation of the tangent at P by letting q tend to p. Find the locus of the midpoint of PQ for chords that pass through the point (2c, 0).

Accepted Answer

The gradient of PQ is (c/q − c/p)/(cq − cp) = (c(p − q)/(pq))/(c(q − p)) = −1/(pq). M1 for the gradient with the factor (p − q) cancelled. The chord is y − c/p = −(1/(pq))(x − cp); multiplying by pq gives pqy − cq = −x + cp, so x + pqy = c(p + q). A1.
As q tends to p the chord turns into the tangent at P: x + p²y = 2cp, the standard form. B1.
The chord passes through (2c, 0) when 2c = c(p + q), that is when p + q = 2. M1. The midpoint of PQ is (c(p + q)/2, c(p + q)/(2pq)) = (c, c/(pq)) for every such chord. A1. Its x-coordinate is always c, so the locus lies on the line x = c; as p runs through the values allowed, pq = p(2 − p) = 1 − (p − 1)² takes every value less than 1 except 0, and never 1 itself, since pq = 1 would need p = q = 1, the degenerate chord with P = Q. So y = c/(pq) covers y > c together with y < 0, and the vertical line is traced with the gap 0 ≤ y ≤ c left out; the point (c, c) would be the midpoint of the tangent at (c, c), not of a chord. A1 for the locus.