Parabola
Pair of Tangents from External Point
Grade 11
Question:
<p>The joint equation of the pair of tangents drawn to the parabola \(y^2 = 4x\) from the point \((-2, -1)\) is:</p>
<p>\(x^2 - 2y^2 + xy + 5x - 2y + 4 = 0\)</p>
<p>\(x^2 + 2y^2 - xy + 5x + 2y - 4 = 0\)</p>
<p>\(x^2 - 2y^2 - xy - 5x + 2y + 4 = 0\)</p>
<p>\(x^2 + 2y^2 + xy - 5x - 2y - 4 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the condition that a line y = mx + c is tangent to y² = 4x when c = 1/m, then apply the constraint that both tangents pass through (-2, -1) to find the pair of slopes. The joint equation is formed by multiplying the two tangent line equations.
<p><strong>Step 1:</strong> For parabola y² = 4x, any tangent line has the form y = mx + 1/m (using c = 1/m where c is the y-intercept).</p><p><strong>Step 2:</strong> Since tangent passes through (-2, -1), substitute: -1 = m(-2) + 1/m, giving -1 = -2m + 1/m.</p><p><strong>Step 3:</strong> Multiply by m: -m = -2m² + 1, or 2m² - m - 1 = 0, which factors as (2m + 1)(m - 1) = 0.</p><p><strong>Step 4:</strong> The two slopes are m₁ = -1/2 and m₂ = 1.</p><p><strong>Step 5:</strong> The tangent lines are:<br>• For m = -1/2: y = -x/2 - 2, or x + 2y + 4 = 0<br>• For m = 1: y = x + 1, or x - y + 1 = 0</p><p><strong>Step 6:</strong> Joint equation = (x + 2y + 4)(x - y + 1) = 0<br>Expanding: x² + 2xy + 4x - xy - 2y² - 4y + x + 2y + 4 = 0<br>= x² + xy - 2y² + 5x - 2y + 4 = 0</p><p>∴ Answer: A</p>
Correct Answer: A