Questions on conic sections

Question 1

Write down parametric coordinates for a general point on the parabola y² = 4ax and on the rectangular hyperbola xy = c².

Accepted Answer

Parabola: (at², 2at). Rectangular hyperbola: (ct, c/t). B1 B1 for the two parametrisations. Both turn a two-variable problem into algebra in a single parameter, and almost every tangent and normal question on this topic starts by using them.

Question 2

For the parabola y² = 20x, state a, the focus and the directrix.

Accepted Answer

4a = 20, so a = 5. The focus is (5, 0) and the directrix is x = −5. B1 for a = 5, B1 for the focus, B1 for the directrix.

Question 3

The ellipse x²/9 + y²/25 = 1 has one semi-axis longer than the other. State the lengths of the two semi-axes and say along which coordinate axis the major axis lies.

Accepted Answer

The denominators are 9 and 25, so the semi-axes are 3 along x and 5 along y. The larger denominator sits under y², so the major axis is vertical, of semi-length 5, and the minor semi-axis is 3. B1 for the two lengths, B1 for the major axis lying along the y-axis. Compare the two denominators before quoting any formula: the larger one is the semi-major axis, whichever letter the equation happens to give it.

Question 4

Write the hyperbola 4x² − y² = 4 in the standard form x²/a² − y²/b² = 1, and write down the coordinates of its vertices and the equations of its asymptotes.

Accepted Answer

Dividing by 4 gives x² − y²/4 = 1, so a = 1 and b = 2. The vertices are (±1, 0). Replacing the 1 by 0 and factorising, x² − y²/4 = 0 gives y = ±2x, so the asymptotes are y = 2x and y = −2x. B1 for the standard form with the vertices, B1 for both asymptotes. The gradient of the asymptotes is b/a, and reading it from the equation before dividing through, as 4/1, gives lines the curve never approaches.

Question 5

For the ellipse x²/16 + y²/7 = 1, find the eccentricity, the foci and the equations of the directrices.

Accepted Answer

a = 4 and b² = 7. From b² = a²(1 − e²): 7 = 16(1 − e²), so e² = 9/16 and e = 3/4. Foci (±ae, 0) = (±3, 0); directrices x = ±a/e = ±16/3. M1 for using b² = a²(1 − e²), A1 for e = 3/4, A1 for the foci, A1 for the directrices. The minus sign inside the bracket is what distinguishes the ellipse formula from the hyperbola's.

Question 6

For the hyperbola x²/16 − y²/9 = 1, find the eccentricity and the foci, and state the difference of focal distances at the vertex (4, 0).

Accepted Answer

Here b² = a²(e² − 1): 9 = 16(e² − 1), so e² = 25/16 and e = 5/4. Foci (±5, 0). From (4, 0) the distances are 1 and 9, differing by 8 = 2a. M1 for using b² = a²(e² − 1), A1 for e = 5/4, A1 for the foci, B1 for the difference 2a. A hyperbola holds the difference of focal distances constant, as an ellipse holds the sum.

Question 7

State the focus-directrix property of a conic, and explain how the eccentricity decides which of the three curves appears.

Accepted Answer

Every point of the conic satisfies distance to the focus = e × distance to the directrix. When e < 1 the curve closes into an ellipse; e = 1 balances it into a parabola; e > 1 opens it into a hyperbola's two branches. B1 for the focus-directrix property, B1 B1 for the three cases of e.

Question 8

The point P(16, 16) lies on the parabola y² = 16x. Verify that it does, and check the focus-directrix property at P.

Accepted Answer

16² = 256 and 16 × 16 = 256, so P is on the curve. Here 4a = 16 and a = 4, so the focus is (4, 0) and the directrix is x = −4. The distance from P to the focus is √(12² + 16²) = 20, and the distance to the directrix is 16 + 4 = 20. Equal, exactly as e = 1 demands. B1 for the verification, B1 for the focus and directrix, M1 for a distance, A1 for both distances equal to 20.

Question 9

For the ellipse x²/9 + y²/25 = 1, find the eccentricity, the coordinates of the foci and the equations of the directrices.

Accepted Answer

The larger denominator is under y², so the major axis is vertical with semi-major axis 5 and semi-minor axis 3. The relation is then 9 = 25(1 − e²), so 1 − e² = 9/25, e² = 16/25 and e = 4/5. The foci lie on the major axis at (0, ±5e) = (0, ±4), and the directrices are the horizontal lines y = ±5/e, that is y = ±25/4. M1 for the eccentricity relation with 25 as the square of the semi-major axis, A1 for e = 4/5, A1 for the foci on the y-axis, A1 for the directrices. Taking 9 as the larger square gives 25 = 9(1 − e²), a negative value of e², which is a signal that the axes have been read the wrong way round, not that the ellipse has none.

Question 10

Sketch the hyperbola x²/9 − y²/4 = 1, drawing its asymptotes first and labelling them with their equations, and marking the vertices. Find the coordinates of the foci and mark them on the sketch.

Accepted Answer

With a = 3 and b = 2 the asymptotes come from x²/9 − y²/4 = 0, so y = ±2x/3. Draw those two lines through the origin, then the vertices at (±3, 0), then the two branches opening left and right, each starting at a vertex and bending towards the lines without crossing them. For the foci, 4 = 9(e² − 1) gives e² = 13/9 and e = √13/3, so ae = √13 and the foci are (±√13, 0), about (±3.61, 0), just outside the vertices on the x-axis. B1 for the asymptotes y = ±2x/3 drawn and labelled, B1 for a sketch with vertices at (±3, 0) and branches approaching the asymptotes, M1 for the eccentricity relation, A1 for the foci (±√13, 0) placed on the sketch. A branch drawn crossing an asymptote, or curling away from it, contradicts what the lines are for: the vertical gap between branch and line shrinks to zero as x grows.

Question 11

A curve has parametric equations x = 3 cos t, y = 2 sin t. Find its Cartesian equation and the coordinates of its foci.

Accepted Answer

cos t = x/3 and sin t = y/2, and cos² t + sin² t = 1 gives x²/9 + y²/4 = 1, an ellipse with a = 3 and b = 2. Then 4 = 9(1 − e²), so e² = 5/9, e = √5/3 and ae = √5: the foci are (±√5, 0). M1 for eliminating t with cos² t + sin² t = 1, A1 for the Cartesian equation, A1 for the foci.

Question 12

The parabola y² = 16x has focus S and directrix d. The tangent at the point P(16, 16) meets d at Q. Find the coordinates of Q and prove that angle PSQ is a right angle.

Accepted Answer

Here a = 4, so S is (4, 0) and d is x = −4. In parametric form P = (at², 2at) gives 2at = 16 with a = 4, so t = 2 and at² = 16 as required. Differentiating y² = 16x implicitly, 2y(dy/dx) = 16, so at P the gradient is 16/32 = 1/2 and the tangent is y − 16 = ½(x − 16), that is 2y = x + 16. Setting x = −4 gives y = 6, so Q(−4, 6). Then SP = (12, 16) and SQ = (−8, 6), and their scalar product is −96 + 96 = 0, so angle PSQ = 90°. B1 for S and d, M1 for differentiating to find the gradient at P, A1 for the tangent 2y = x + 16, M1 for putting x = −4, A1 for Q(−4, 6), M1 for the scalar product of SP and SQ, A1 for the conclusion. This holds for every point of every parabola, not only for P; repeating the working with (at², 2at) gives SP·SQ = 0 in general.

Question 13

A point P(x, y) moves so that its distance from S(4, 0) is always twice its distance from the line x = 1. Show that the locus of P is a hyperbola, find its equation in the form x²/a² − y²/b² = 1, and state its eccentricity and the equations of its asymptotes.

Accepted Answer

The condition is SP = 2 × (distance to x = 1), so √((x − 4)² + y²) = 2|x − 1|. Squaring, (x − 4)² + y² = 4(x − 1)², that is x² − 8x + 16 + y² = 4x² − 8x + 4. The x terms cancel, leaving 3x² − y² = 12, so x²/4 − y²/12 = 1. This is a hyperbola with a = 2 and b² = 12, since a point is e = 2 times as far from a fixed point as from a fixed line and e > 1. Check: b² = a²(e² − 1) gives 12 = 4(e² − 1), so e² = 4 and e = 2, with focus (ae, 0) = (4, 0) and directrix x = a/e = 1, the data we started from. The asymptotes are y = ±√3 x, from x²/4 − y²/12 = 0. M1 for writing the distance condition, A1 for the squared equation, M1 for expanding and collecting, A1 for x²/4 − y²/12 = 1, B1 for identifying a hyperbola with e = 2 from the focus-directrix definition or from the b² relation, B1 for the asymptotes. The distance to the line is |x − 1|, and squaring removes the modulus; writing (x − 1) without it is harmless here only because the squaring follows immediately.

Question 14

The hyperbola x²/a² − y²/b² = 1 has eccentricity e, with b² = a²(e² − 1), and S is the focus (ae, 0). Show that the point P(a sec t, b tan t) lies on the hyperbola for every value of t, and prove that for a point P on the right-hand branch SP = a(e sec t − 1).

Accepted Answer

Substituting, (a sec t)²/a² − (b tan t)²/b² = sec² t − tan² t = 1, so P is on the curve for every t. B1 for that. Then SP² = (a sec t − ae)² + b² tan² t. M1 for the distance formula in terms of t. Replace b² by a²(e² − 1) and tan² t by sec² t − 1: SP² = a² sec² t − 2a²e sec t + a²e² + a²(e² − 1)(sec² t − 1). M1 for both substitutions. Multiplying out the last bracket gives a²e² sec² t − a²e² − a² sec² t + a², and adding it to the first three terms leaves SP² = a²e² sec² t − 2a²e sec t + a² = a²(e sec t − 1)². A1 for the collected square. Taking the square root, SP = a|e sec t − 1|, and on the right-hand branch sec t is at least 1 while e > 1, so e sec t > 1 and SP = a(e sec t − 1). A1 for the result, B1 for the sign argument that chooses the positive root. As a check, the distance from P to the directrix x = a/e is a sec t − a/e = (a/e)(e sec t − 1), and SP is e times it, which is the focus-directrix property. On the left-hand branch sec t is at most −1 and the same working gives SP = a(1 − e sec t).

Question 15

A point P(x, y) moves so that the sum of its distances from A(0, 3) and B(0, −3) is always 10. Show that the locus of P is an ellipse, find its Cartesian equation, and state its eccentricity and the direction of its major axis.

Accepted Answer

Write the condition as √(x² + (y − 3)²) = 10 − √(x² + (y + 3)²). M1 for isolating one root. Squaring, x² + (y − 3)² = 100 − 20√(x² + (y + 3)²) + x² + (y + 3)². The x² terms cancel and (y − 3)² − (y + 3)² = −12y, so −12y = 100 − 20√(x² + (y + 3)²), that is 5√(x² + (y + 3)²) = 25 + 3y. A1 for this line. Squaring again, 25(x² + y² + 6y + 9) = 625 + 150y + 9y². M1 for the second squaring. The 150y terms cancel: 25x² + 16y² = 400, so x²/16 + y²/25 = 1. A1. The larger denominator is under y², so this is an ellipse with its major axis vertical, semi-major axis 5 and semi-minor axis 4. B1 for identifying the ellipse and the direction of its major axis. From 16 = 25(1 − e²), e² = 9/25 and e = 3/5, so the foci are (0, ±5e) = (0, ±3), which are A and B themselves. B1 for e = 3/5. The constant sum 10 is 2b, the length of the major axis, and 3 = be is the focal distance, exactly as the ellipse's definition by two foci demands.