3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
For a real number k, consider the curve 10−kx2+k−4y2=1. The set of all values of k for which this curve is an ellipse (not a circle) whose major axis lies along the y-axis and whose two foci are at a distance of at most 4 from each other is ______. Exactly one option is correct.
A(7,10)
B(7,9]
C[7,9]
D(4,9]
Show answer and solution
Answer: B
Write A=10−k, B=k−4. An ellipse needs A>0, B>0; the major axis is along the y-axis and the curve is not a circle exactly when B>A>0, i.e. k−4>10−k and k<10, so 7<k<10.
The semi-major axis is B, the semi-minor axis A, so the focal distance is 2B−A=22k−14. The condition 22k−14≤4 gives 2k−14≤4, i.e. k≤9.
Hence k∈(7,9]. At k=7 the curve is the circle x2+y2=3, excluded; at k=9 the foci are (0,±2), exactly 4 apart, included.
Trap: Including k=7 (a circle, whose 'foci' coincide at the centre) gives the option [7,9].
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
An ellipse has a focus at S(2,1), the corresponding directrix x+y+1=0 and eccentricity 21. The length of its major axis is ______. Exactly one option is correct.
A342
B38
C3162
D382
Show answer and solution
Answer: D
For any ellipse, the distance from a focus to its directrix is ea−ae=ea(1−e2).
Here that distance is 2∣2+1+1∣=22, and e=21 gives 21a⋅43=23a=22, so a=342 and the major axis is 2a=382.
Check directly: the major axis is the line through S perpendicular to the directrix, P=S+2t(1,1). Then SP=∣t∣ and the distance from P to the directrix is ∣t+22∣. Solving ∣t∣=21∣t+22∣ gives the vertices at t=22 and t=−322, which are 382 apart.
Difficulty: Hard
Question 3Level 2 · ExamSingle correct
For the ellipse 9x2+25y2−18x+100y−116=0, the greatest possible distance of its centre from a normal to the ellipse is ______. Exactly one option is correct.
A2
B516
C4
D1
Show answer and solution
Answer: A
Completing squares: 9(x−1)2+25(y+2)2=116+9+100=225, i.e. 25(x−1)2+9(y+2)2=1: centre C(1,−2), a=5, b=3.
Shift the origin to C. The normal at (5cosθ,3sinθ) is 5xsecθ−3ycscθ=16 (for sinθcosθ=0; at the vertices the normals pass through C). Its distance from C is
d=25sec2θ+9csc2θ16=34+25tan2θ+9cot2θ16≤34+2⋅1516=816=2,
with equality when tan2θ=53. So the maximum is a−b=2.
Difficulty: Hard
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