Hyperbola
Tangents to hyperbola from external point
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: A tangent to a hyperbola from an external point touches one branch. For two tangents to touch different branches, the point must lie in the region where two distinct tangent lines can be drawn—one to each branch. This requires analyzing the condition on α using the tangent equation and discriminant.
<p><strong>Step 1:</strong> The hyperbola is <strong>x² - y²/4 = 1</strong>. A tangent line at slope m is: <strong>y = mx ± √(1 - 4m²)</strong></p><p><strong>Step 2:</strong> For the point (α, α²) to lie on a tangent: <strong>α² = mα ± √(1 - 4m²)</strong></p><p><strong>Step 3:</strong> Rearranging: <strong>√(1 - 4m²) = ±(α² - mα)</strong></p><p>Squaring: <strong>1 - 4m² = α⁴ - 2α³m + α²m²</strong></p><p><strong>Step 4:</strong> Rearranging as quadratic in m: <strong>(α² + 4)m² - 2α³m + (α⁴ - 1) = 0</strong></p><p><strong>Step 5:</strong> For two real tangents: Δ > 0</p><p><strong>Δ = 4α⁶ - 4(α² + 4)(α⁴ - 1) > 0</strong></p><p><strong>Δ = 4α⁶ - 4(α⁶ - α² + 4α⁴ - 4) > 0</strong></p><p><strong>Δ = 4(-4α⁴ + α² + 4) > 0</strong></p><p><strong>4α⁴ - α² - 4 < 0</strong></p><p><strong>Step 6:</strong> The two tangents touch different branches when the point lies between the branches, requiring: <strong>(4α⁴ - α² - 4) < 0</strong> which gives <strong>|α| < 1</strong> (approximately, depending on precise roots of 4t² - t - 4 = 0)</p><p><strong>Step 7:</strong> Solving 4t² - t - 4 = 0 where t = α²: <strong>t = (1 ± √(1+64))/8 = (1 ± √65)/8</strong></p><p>Since t = α² ≥ 0, we need <strong>0 ≤ α² < (1 + √65)/8</strong> or equivalently <strong>|α| < √[(1+√65)/8]</strong></p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C