Hyperbola
Tangents to hyperbola
Grade 11
Question:
<p>If two tangents can be drawn to the different branches of the hyperbola \(x^2 - \dfrac{y^2}{4} = 1\) from the point \((\alpha,\, \alpha^2)\), then:</p>
<p>\(\alpha \in (-\infty,\,-3)\)</p>
<p>\(\alpha \in (3,\,\infty)\)</p>
<p>\(\alpha \in (-2,\,0) \cup (0,\,2)\)</p>
<p>\(a \in (2,\,\infty)\)</p>
Step-by-Step Solution
Key Concept: For tangents from an external point to different branches of a hyperbola, the point must lie outside the hyperbola AND the chord of contact must intersect both branches. The critical condition is that the point lies in the region where tangents to different branches exist.
<p><strong>Step 1:</strong> For tangents from point (α, α²) to the hyperbola x² - y²/4 = 1, substitute into the tangent equation x·x₀ - (y·y₀)/4 = 1, giving: αx - (α²·y)/4 = 1.</p><p><strong>Step 2:</strong> For two tangents to exist from (α, α²), this line must intersect the hyperbola in two real points. This requires the discriminant condition to be satisfied.</p><p><strong>Step 3:</strong> For tangents to DIFFERENT branches specifically, the chord of contact must cut both branches. This means the point (α, α²) must lie between the asymptotes: |α²| < 4α² is automatically satisfied, but we need α² < α⁴ - 1, which gives α⁴ - α² - 1 > 0.</p><p><strong>Step 4:</strong> Solving α⁴ - α² - 1 > 0: Let u = α², then u² - u - 1 > 0. This gives u > (1+√5)/2, so α² > (1+√5)/2.</p><p><strong>Step 5:</strong> Additionally, for the geometry to work (tangents to different branches from an internal region), we need |α| > √[(1+√5)/2] ≈ 1.27, which restricts α to approximately |α| > 1.27.</p><p>∴ Answer: A</p>
Correct Answer: A