A point \( P \) in the first quadrant, lies on \( y^2 = 4ax \), \( a > 0 \), and keeps a distance of \( 5a \) units from its focus. Which of the following points lies on the locus of \( P \)?
Show Hint
- \( (1, 0) \)
- \( (1, 1) \)
- \( (0, 2) \)
- \( (2, 0) \)
The Correct Option is B
Solution and Explanation
Step 1: General form of the parabola.
The equation of the given parabola is \( y^2 = 4ax \), which represents a parabola opening to the right with vertex at the origin. The focus of this parabola is at \( (a, 0) \), and the distance from any point on the parabola to the focus is equal to its distance from the directrix.
Step 2: Condition for the locus of \( P \).
We are told that the point \( P \) keeps a distance of \( 5a \) units from the focus. This means that the distance from \( P(x, y) \) to the focus \( (a, 0) \) is \( 5a \). The distance between any point \( P(x, y) \) and the focus \( (a, 0) \) is given by:
\[
\text{Distance from } P \text{ to focus} = \sqrt{(x - a)^2 + y^2}.
\]
Since this distance is \( 5a \), we have the equation:
\[
\sqrt{(x - a)^2 + y^2} = 5a.
\]
Squaring both sides:
\[
(x - a)^2 + y^2 = 25a^2.
\]
Step 3: Substitute the equation of the parabola.
We know from the equation of the parabola that \( y^2 = 4ax \). Substituting this into the equation above:
\[
(x - a)^2 + 4ax = 25a^2.
\]
Expanding and simplifying:
\[
x^2 - 2ax + a^2 + 4ax = 25a^2,
\]
\[
x^2 + 2ax + a^2 = 25a^2,
\]
\[
x^2 + 2ax - 24a^2 = 0.
\]
Step 4: Solve for \( x \) and \( y \).
Now, we substitute the options for \( x \) and \( y \) and check which one satisfies the equation.
- For \( (1, 1) \), substitute \( x = 1 \) and \( y = 1 \):
\[
1^2 + 2a(1) - 24a^2 = 0,
\]
\[
1 + 2a - 24a^2 = 0.
\]
Solving for \( a \), we find that this satisfies the equation. Hence, the point \( (1, 1) \) lies on the locus of \( P \).
Step 5: Conclusion.
Thus, the correct answer is \( (1, 1) \), and the correct option is (b).
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