Question

The axis of a parabola is the line $ y = x $ and its vertex and focus are in the first quadrant at distances $ \sqrt{2} $ and $ 2\sqrt{2} $ units from the origin, respectively. If the point $ (1, k) $ lies on the parabola, then a possible value of $ k $ is:

• A. 4
• B. 9
• C. 3
• D. 8

Hint:

When solving problems with parabolas, use the definition of a parabola that equates the distances from any point on the curve to the focus and the directrix.

Answer 1

The Correct Option is B

The parabola's vertex is at \( (0, 0) \), and its axis is the line \( y = x \). The focus is at \( (2\sqrt{2}, 2\sqrt{2}) \), and the directrix is \( x + y = 0 \).
By the definition of a parabola, any point on it is equidistant from the focus and the directrix. Consider the point \( P(1, k) \) on the parabola. 
Let \( PS \) be the distance from \( P \) to the focus, and \( PM \) be the distance from \( P \) to the directrix. 
The distance \( PS \) is calculated as: \[ PS = \sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2} \] The distance \( PM \) from \( P(1, k) \) to the directrix \( x + y = 0 \) is: \[ PM = \frac{|1 \times 1 + k|}{\sqrt{1^2 + 1^2}} = \frac{|1 + k|}{\sqrt{2}} \] Equating \( PS \) and \( PM \) yields: \[ \sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2} = \frac{|1 + k|}{\sqrt{2}} \] Solving this equation gives \( k = 9 \). 
Therefore, the answer is \( 9 \).