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).
Let \( F_1, F_2 \) \(\text{ be the foci of the hyperbola}\) \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, a > 0, \, b > 0, \] and let \( O \) be the origin. Let \( M \) be an arbitrary point on curve \( C \) and above the X-axis and \( H \) be a point on \( MF_1 \) such that \( MF_2 \perp F_1 F_2, \, M F_1 \perp OH, \, |OH| = \lambda |O F_2| \) with \( \lambda \in (2/5, 3/5) \), then the range of the eccentricity \( e \) is in:
Let the line $\frac{x}{4} + \frac{y}{2} = 1$ meet the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line $x + y = 1$. Let the point P lie on the line $x + y = 1$ such that $\Delta ABP$ is an isosceles triangle with $AP = BP$. Then the distance between M' and P is:
A remote island has a unique social structure. Individuals are either "Truth-tellers" (who always speak the truth) or "Tricksters" (who always lie). You encounter three inhabitants: X, Y, and Z.
X says: "Y is a Trickster"
Y says: "Exactly one of us is a Truth-teller."
What can you definitively conclude about Z?
Consider the following statements followed by two conclusions.
Statements: 1. Some men are great. 2. Some men are wise.
Conclusions: 1. Men are either great or wise. 2. Some men are neither great nor wise. Choose the correct option: