If $x - 2y + k = 0$ is a tangent to the parabola $y^2 - 4x - 4y + 8 = 0$, then the value of $k$ is:
2
\(\frac{2}{5}\)
7
-7
The Correct Option is C
Solution and Explanation
To find the value of \(k\) such that the line \(x - 2y + k = 0\) is tangent to the parabola \(y^2 - 4x - 4y + 8 = 0\), we proceed step by step.
Step 1: Convert parabola to standard form
\[ y^2 - 4x - 4y + 8 = 0 \]
\[ y^2 - 4y = 4x - 8 \]
Complete the square:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
\[ (y - 2)^2 = 4(x - 1) \]
The parabola has vertex \( (1,2) \).
Step 2: Tangency condition
For tangency, the line \(x - 2y + k = 0\) must satisfy the condition using substitution.
Substitute the vertex into the line expression:
\[ d = \frac{|1 - 2(2) + k|}{\sqrt{1^2 + (-2)^2}} = \frac{|k - 3|}{\sqrt{5}} \]
For tangency with this parabola, the condition becomes:
\[ \frac{|k - 3|}{\sqrt{5}} = 2 \]
Step 3: Solve
\[ |k - 3| = 2\sqrt{5} \]
So:
\[ k = 3 \pm 2\sqrt{5} \]
Final observation: both values are irrational, so there is no integer value of \(k\) that satisfies exact tangency.
Final Answer: \(3 \pm 2\sqrt{5}\)
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Concepts Used:
Coordinate Geometry
The central idea in coordinate geometry is to assign numerical coordinates to points in a plane or space, which allows us to describe their positions and relationships using algebraic equations. The most common coordinate system is the Cartesian coordinate system, named after the French mathematician and philosopher René Descartes.