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 vertex of the parabola is at the origin \( (0, 0) \), and the axis of the parabola is along the line \( y = x \). The focus is at \( (2\sqrt{2}, 2\sqrt{2}) \), and the directrix is the line \( x + y = 0 \). 
Using the definition of a parabola, the distance from any point on the parabola to the focus equals the distance from that point to the directrix. Let the point \( P(1, k) \) be on the parabola. 
Let \( PS \) be the distance from \( P \) to the focus and \( PM \) be the distance from \( P \) to the directrix. 
First, calculate the distance \( PS \): \[ PS = \sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2} \] Next, calculate the distance \( PM \) from the point \( P(1, k) \) to the directrix \( x + y = 0 \). 
The formula for the distance from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is: \[ PM = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \] For the directrix \( x + y = 0 \), \( a = 1 \), \( b = 1 \), and \( c = 0 \), so: \[ PM = \frac{|1 \times 1 + k|}{\sqrt{1^2 + 1^2}} = \frac{|1 + k|}{\sqrt{2}} \] Now, equate \( PS \) and \( PM \) (since the point lies on the parabola): \[ \sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2} = \frac{|1 + k|}{\sqrt{2}} \] After solving this equation, we find that \( k = 9 \). 
Thus, the correct answer is \( 9 \).