Let O(0, 0) and A(0, 1) be two fixed points. Then the locus of a point P such that the perimeter of

Q: Wait, why is the sum of the distances equal to 3 and not 4?

Great catch! The perimeter of the triangle is the sum of all three sides: $OA + OP + AP = 4$. Since $O$ and $A$ are fixed points, the distance $OA$ is constant. Calculating the distance between $(0,0)$ and $(0,1)$ gives us 1. Subtracting this from the total perimeter of 4 leaves us with $OP + AP = 3$. Always remember to subtract the fixed segment from the total perimeter!

Q: Why can't we just use the standard ellipse formula $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$?

You absolutely could, but only after shifting the center to the origin! The standard formula assumes the center of the ellipse is at $(0,0)$. Here, our foci are at $(0,0)$ and $(0,1)$, meaning the center is at $(0, 0.5)$. Using the distance formula directly is often safer and less prone to errors regarding coordinate shifts.

Q: Is it possible for the locus to be a hyperbola instead?

Excellent question! A hyperbola is defined by the *difference* of distances being constant ($|PF_1 - PF_2| = 2a$). Since our problem specifies the *sum* of distances is constant, we are strictly in the realm of the ellipse. The sum constraint keeps the point 'trapped' between the foci, whereas a difference constraint would push it away.

Q: I keep getting a mess when I square the equation. Is there a trick?

The trick is to isolate one radical completely before squaring. If you square $\sqrt{A} + \sqrt{B} = C$ directly, you get $A + B + 2\sqrt{AB} = C^2$, which still leaves you with a radical. By moving one term first, you ensure that the second squaring step will eliminate the remaining radical entirely.

Q: How do I know if my final equation is correct?

Check the coefficients! For an ellipse, the coefficients of $x^2$ and $y^2$ must have the same sign. If you end up with different signs, you've accidentally wandered into hyperbola territory. Also, test a simple point, like the midpoint of the foci, to see if it satisfies your final equation.

Question

Let $O(0, 0)$ and $A(0, 1)$ be two fixed points. Then the locus of a point $P$ such that the perimeter of $\Delta AOP$ is 4, is :

Options:

(A) $8x^2 - 9y^2 + 9y = 18$

(B) $9x^2 + 8y^2 - 8y = 16$

(C) $8x^2 + 9y^2 - 9y = 18$

(D) $9x^2 - 8y^2 + 8y = 16$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plane. You have two stakes driven into the ground at points $O(0, 0)$ and $A(0, 1)$.

You are holding a string, and you want to walk in such a way that the total perimeter of the triangle formed by you and these two stakes remains exactly $4$ units long. You are tracing one of the most beautiful shapes in mathematics: the ellipse.

### The Hidden Constraint

Let us look at our triangle $\Delta AOP$. The perimeter is defined as the sum of its three sides: $OA + OP + AP = 4$.

We know the coordinates of our fixed points $O(0, 0)$ and $A(0, 1)$. Using the distance formula, the length of the segment $OA$ is simply:

$$OA = \sqrt{(0-0)^2 + (1-0)^2} = 1$$

Now, substitute this into our perimeter equation: $1 + OP + AP = 4$. This simplifies to $OP + AP = 3$.

This is the heartbeat of our problem. We have a moving point $P(x, y)$ whose total distance from two fixed points is a constant value of $3$. In the language of geometry, $O$ and $A$ are the foci of an ellipse, and the constant sum $3$ represents the length of the major axis, $2a$.

### The Algebraic Dance

To find the equation of this path, we translate our geometric intuition into algebra. The distance $OP$ is $\sqrt{x^2 + y^2}$, and the distance $AP$ is $\sqrt{x^2 + (y-1)^2}$.

Our condition is:

$$\sqrt{x^2 + y^2} + \sqrt{x^2 + (y-1)^2} = 3$$

The secret to handling these radicals is to isolate one of them. Let us move the $OP$ term to the right side:

$$\sqrt{x^2 + (y-1)^2} = 3 - \sqrt{x^2 + y^2}$$

Now, we square both sides. On the left, the radical vanishes. On the right, we use the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$$x^2 + (y-1)^2 = 9 + (x^2 + y^2) - 6\sqrt{x^2 + y^2}$$

### The Elegant Cancellation

Expand the term $(y-1)^2$ to get $y^2 - 2y + 1$. Our equation now looks like this:

$$x^2 + y^2 - 2y + 1 = 9 + x^2 + y^2 - 6\sqrt{x^2 + y^2}$$

Look closely—the $x^2$ and $y^2$ terms appear on both sides. They cancel out perfectly, leaving us with a much simpler expression:

$$-2y + 1 = 9 - 6\sqrt{x^2 + y^2}$$

Rearranging to isolate the remaining radical, we get $-2y - 8 = -6\sqrt{x^2 + y^2}$. Dividing everything by $-2$ makes the numbers manageable:

$$y + 4 = 3\sqrt{x^2 + y^2}$$

### The Final Reveal

We are almost there. One final square to remove the last radical:

$$(y + 4)^2 = (3\sqrt{x^2 + y^2})^2$$

$$y^2 + 8y + 16 = 9(x^2 + y^2)$$

Expanding the right side gives $9x^2 + 9y^2$. Bringing everything to one side, we arrive at the final locus:

$$9x^2 + 8y^2 - 8y = 16$$

This is the equation of our ellipse. It is the mathematical signature of every point $P$ that keeps the perimeter of $\Delta AOP$ constant. You have successfully navigated the geometry, the algebra, and the logic to reach this **elegant result**.

Animated Solution for Mathematics - Conic Sections: Let $O (0, 0)$ and $A (0, 1)$ be two fixed points. Then the locus of a point $P$ such that the perimeter of $Δ A O P$ is 4, is :

Select Answer:

Visualized Solution

Visualize the Fixed Points $O$ and $A$

Define the Moving Point $P (x, y)$

Calculate the Sum of Distances

Recognize the Geometric Shape

Apply the Distance Formula

Isolate One Radical Term

Square Both Sides

Simplify and Cancel Terms

Isolate the Remaining Radical

Square Both Sides Again

Final Algebraic Simplification

Conclusion and Final Result

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Hidden Constraint

The Algebraic Dance

The Elegant Cancellation

The Final Reveal

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let O(0,0) and A(0,1) be two fixed points. Then the locus of a point P such that the perimeter of ΔAOP is 4, is :