Key Concepts and Formulas

• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Definition of an Ellipse: The locus of a point such that the sum of its distances from two fixed points (called foci) is a constant, is an ellipse.

Step-by-Step Solution

Let the coordinates of the variable point be $P(x, y)$. The fixed points are $O(0, 0)$ and $A(0, 1)$.

Step 1: Understand the Given Condition The perimeter of $\Delta AOP$ is given as 4. The perimeter of a triangle is the sum of the lengths of its sides. So, $OA + OP + AP = 4$.

Step 2: Calculate the Distance Between the Fixed Points Let's find the length of the side $OA$. Using the distance formula for $O(0, 0)$ and $A(0, 1)$: $OA = \sqrt{(0-0)^2 + (1-0)^2} = \sqrt{0^2 + 1^2} = \sqrt{1} = 1$

Step 3: Simplify the Perimeter Condition Substitute the value of $OA$ into the perimeter equation: $1 + OP + AP = 4$ This simplifies to: $OP + AP = 3$ This is the defining condition for the locus of P.

Step 4: Express Distances OP and AP in terms of $x$ and $y$ Using the distance formula for $P(x, y)$ and $O(0, 0)$: $OP = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$ Using the distance formula for $P(x, y)$ and $A(0, 1)$: $AP = \sqrt{(x-0)^2 + (y-1)^2} = \sqrt{x^2 + (y-1)^2}$

Now, substitute these expressions back into the simplified locus condition $OP + AP = 3$: $\sqrt{x^2 + y^2} + \sqrt{x^2 + (y-1)^2} = 3$

Step 5: Eliminate the Square Roots (First Time) To eliminate square roots, isolate one radical term and then square both sides. Let's isolate $\sqrt{x^2 + (y-1)^2}$: $\sqrt{x^2 + (y-1)^2} = 3 - \sqrt{x^2 + y^2}$ Now, square both sides of the equation: $(\sqrt{x^2 + (y-1)^2})^2 = (3 - \sqrt{x^2 + y^2})^2$ $x^2 + (y-1)^2 = 3^2 - 2(3)\sqrt{x^2 + y^2} + (\sqrt{x^2 + y^2})^2$ $x^2 + y^2 - 2y + 1 = 9 - 6\sqrt{x^2 + y^2} + x^2 + y^2$

Step 6: Simplify and Isolate the Remaining Square Root Notice that $x^2 + y^2$ appears on both sides of the equation. We can cancel these terms: $x^2 + y^2 - 2y + 1 = 9 - 6\sqrt{x^2 + y^2} + x^2 + y^2$ $-2y + 1 = 9 - 6\sqrt{x^2 + y^2}$ Now, isolate the remaining square root term, $6\sqrt{x^2 + y^2}$. Move it to the LHS and other terms to the RHS: $6\sqrt{x^2 + y^2} = 9 - 1 + 2y$ $6\sqrt{x^2 + y^2} = 8 + 2y$ We can divide the entire equation by 2 to simplify the coefficients: $3\sqrt{x^2 + y^2} = 4 + y$

Step 7: Eliminate the Square Root (Second Time) To remove the last square root, we square both sides of the equation again: $(3\sqrt{x^2 + y^2})^2 = (4 + y)^2$ $3^2 (x^2 + y^2) = 4^2 + 2(4)(y) + y^2$ $9(x^2 + y^2) = 16 + 8y + y^2$

Step 8: Rearrange into the Standard Form of the Locus Equation Distribute the 9 on the LHS: $9x^2 + 9y^2 = 16 + 8y + y^2$ Now, move all terms to one side to get the standard form of the equation: $9x^2 + 9y^2 - y^2 - 8y - 16 = 0$ $9x^2 + 8y^2 - 8y - 16 = 0$ This can also be written as: $9x^2 + 8y^2 - 8y = 16$

This is the equation of the locus of point P.

Common Mistakes & Tips

• Algebraic Precision: Be extremely careful with algebraic manipulations, especially when squaring terms involving sums or differences.
• Isolate Radicals: When dealing with equations involving multiple square roots, always isolate one square root term before squaring.

Summary

The problem asked for the locus of a point P such that the perimeter of $\Delta AOP$ is 4, with fixed points $O(0,0)$ and $A(0,1)$. Through careful algebraic manipulation involving two rounds of squaring to eliminate square roots, we arrived at the equation of the ellipse: $9x^2 + 8y^2 - 8y = 16$. This matches option (A).

The final answer is \boxed{9x^2 + 8y^2 - 8y = 16}, which corresponds to option (A).