Director Circle Practice
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Direct answer
From any point on the circle x² + y² = a² + b², the two tangents drawn to the ellipse x²/a² + y²/b² = 1 are perpendicular — that circle is the director circle, the locus where right-angled tangent pairs live. For the hyperbola x²/a² − y²/b² = 1 the locus is x² + y² = a² − b², real only when a > b; the rectangular hyperbola (a = b) collapses it to the single point (0, 0). The parabola has no director circle at all: perpendicular tangents to y² = 4ax meet on its directrix instead. One sign — plus b² or minus b² — separates the ellipse's answer from the hyperbola's.
What you must remember
- Ellipse: director circle x² + y² = a² + b², concentric with the ellipse and larger than the auxiliary circle since a² + b² > a².
- Hyperbola: x² + y² = a² − b², real only for a > b; the equilateral case a = b degenerates to the origin.
- Parabola: no director circle — tangents at the ends of any focal chord (t₁t₂ = −1) are perpendicular and meet on the directrix x = −a.
- Slope-form engine: the ellipse's tangent y = mx ± √(a²m² + b²) with the perpendicularity condition m₁m₂ = −1 generates the locus by elimination.
- Angle reading: from a point on the director circle the tangent-contact angle is exactly 90°; closer in it is obtuse, farther out acute — a counting-tool for angle questions.
- Hyperbola slope form: tangent y = mx ± √(a²m² − b²), needing |m| ≥ b/a; the same elimination yields a² − b².
- Degenerate flag: when a² − b² < 0 no real director circle exists for the hyperbola — a true/false staple.
Deriving the director circle
Work the ellipse through the quadratic-in-m trick. A line through (h, k) with slope m is tangent to x²/a² + y²/b² = 1 exactly when its intercept satisfies c² = a²m² + b² with c = k − mh; expanding gives m²(h² − a²) − 2hkm + (k² − b²) = 0. The two tangents from (h, k) have slopes m₁ and m₂, the two roots of this quadratic, so Vieta's product gives m₁m₂ = (k² − b²)/(h² − a²). Perpendicularity forces that product to be −1: k² − b² = −(h² − a²), hence h² + k² = a² + b². The locus is the director circle, derived in four lines.
The craft is the reformulation: writing "two tangents from a point" as "two roots of a quadratic in m" converts a geometry problem into Vieta's formulas. Run the identical three lines with c² = a²m² − b² for the hyperbola and the product of roots becomes (k² − b²)/(h² − a²) again but with the condition landing on a² − b² — the sign difference between the two director circles traces back to a single minus in the tangent condition. The same quadratic-in-m skeleton also handles chord-of-contact and normal-count questions, which is why it deserves over memorising the two locus equations.
Ellipse versus hyperbola
Main asks the locus directly — "locus of the point of intersection of perpendicular tangents" — with options mixing a² + b² against a² − b²; Advanced wraps the director circle into tangent-counting or angle-between-tangents items and probes the rectangular hyperbola's degeneracy. The traps: quoting a² + b² for the hyperbola (the sign); asserting a real director circle for every hyperbola (a > b is required); awarding the parabola one (it has none — the directrix takes over the role); and forgetting that a point on the ellipse itself admits only one tangent, so the locus necessarily sits strictly outside the conic. When a numerical question hides the director circle — "the tangents from (h, k) to the ellipse are perpendicular, find k² + h²" — the answer is the constant a² + b² without any coordinate work.
Frequently asked questions
What is the director circle of an ellipse?
The circle x² + y² = a² + b² — the locus of points from which the two tangents to the ellipse meet at right angles.
Why does the rectangular hyperbola have a point director circle?
a = b makes a² − b² = 0, so the locus x² + y² = 0 collapses to the origin: perpendicular tangent pairs meet only at the centre.
Does a parabola have a director circle?
No — perpendicular tangents to a parabola meet on its directrix, which plays the director-circle role instead.
How does the slope form derive the director circle?
Write the tangent condition from (h, k) as a quadratic in m; the product of roots (k² − b²)/(h² − a²) must equal −1, giving h² + k² = a² + b².
Where do perpendicular tangents to y² = 4ax meet?
On the directrix x = −a: the ends of any focal chord (t₁t₂ = −1) carry perpendicular tangents meeting at (−a, a(t₁ + t₂)).