Hyperbola
Tangent Lines
Grade 11
Question:
<p>The circle \(x^2 + y^2 - 8x = 0\) and hyperbola \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) intersect at the points A and B. Find the equation of a common tangent with positive slope to the circle as well as to the hyperbola.</p>
<p>(a) \(2x - 5y - 20 = 0\)</p>
<p>(b) \(2x - 5y + 4 = 0\)</p>
<p>(c) \(3x - 4y + 8 = 0\)</p>
<p>(d) \(4x - 3y + 4 = 0\)</p>
Step-by-Step Solution
Key Concept: A common tangent must satisfy both the tangency condition for the circle (perpendicular distance = radius) and the tangency condition for the hyperbola (\(c^2 = a^2m^2 - b^2\)).
<p>For a line \(y = mx + c\) to be tangent to the circle \(x^2 + y^2 - 8x = 0\) (centre (4,0), radius 4), the perpendicular distance from the centre to the line must equal the radius. For tangency to the hyperbola \(\frac{x^2}{9} - \frac{y^2}{4} = 1\), the condition \(c^2 = 9m^2 - 4\) must hold. Solving these conditions with positive slope gives \(2x - 5y + 4 = 0\).</p>
Correct Answer: B