Let P and Q be distinct points on the parabola y² = 2x such that a circle with PQ as diameter passes

Q: Wait, why does the circle passing through the origin mean $\angle POQ = 90^\circ$?

This is a classic property from circle geometry! Any angle subtended by a diameter at any point on the circumference of a circle is always a right angle. Since $PQ$ is the diameter and $O$ lies on the circle, the triangle $\Delta POQ$ must be a right-angled triangle at $O$.

Q: Why did we use $t_1$ and $t_2$ instead of just solving for $x$ and $y$ directly?

Great question! Solving for $x$ and $y$ directly leads to messy algebraic equations involving square roots. By using the parametric form $P = (\frac{t^2}{2}, t)$, we reduce the problem to finding the values of the parameters $t_1$ and $t_2$, which turns a complex geometry problem into a simple quadratic equation.

Q: How do I know if I should take the positive or negative root for $t_1$?

Always look back at the problem statement! We are told that $P$ lies in the first quadrant. Since the y-coordinate of $P$ is $t_1$, this forces $t_1$ to be positive. If you get a negative value for $t_1$, it simply means that point corresponds to a location in the fourth quadrant, which we must discard.

Q: Is it possible for $t_1$ and $t_2$ to be the same?

The problem explicitly states that $P$ and $Q$ are distinct points. If $t_1 = t_2$, then $P$ and $Q$ would be the same point, and we wouldn't have a triangle at all! Our condition $t_1 t_2 = -4$ ensures they are distinct.

Q: Why did we use the identity $(t_1 + t_2)^2 = (t_1 - t_2)^2 + 4t_1 t_2$?

We found the value of $|t_1 - t_2|$ from the area, and we knew the product $t_1 t_2$ from the perpendicularity condition. This identity is the bridge that allows us to find the sum $(t_1 + t_2)$, which is exactly what we need to form the quadratic equation for $t$.

Question

Let $P$ and $Q$ be distinct points on the parabola $y^2 = 2x$ such that a circle with $PQ$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $\Delta OPQ$ is $3\sqrt{2}$, then which of the following is (are) the coordinates of $P$?

Options:

(A) $(4, 2\sqrt{2})$

(B) $(9, 3\sqrt{2})$

(C) $(\frac{1}{4}, \frac{1}{\sqrt{2}})$

(D) $(1, \sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the curve $y^2 = 2x$. It is a beautiful, sweeping parabola, opening wide to the right.

We have two points, $P$ and $Q$, dancing on this curve. A circle drawn with the segment $PQ$ as its diameter passes through the origin $O(0,0)$.

### The Power of Parametric Elegance

When dealing with parabolas, the Cartesian coordinates $(x, y)$ can sometimes feel like a heavy cloak. To move freely, we use the parametric form.

For $y^2 = 2x$, we identify $4a = 2$, which gives us $a = \frac{1}{2}$. Thus, any point on this parabola can be represented as:

$$P = \left( \frac{t_1^2}{2}, t_1 ight), \quad Q = \left( \frac{t_2^2}{2}, t_2 ight)$$

By shifting our perspective to these parameters $t_1$ and $t_2$, we transform a daunting coordinate geometry problem into a clean, algebraic one.

### The Orthogonality Insight

If a circle with diameter $PQ$ passes through the origin $O$, then the angle $\angle POQ$ must be $90^{\circ}$. This is because the angle subtended by a diameter at any point on the circle is always a right angle.

This implies that the line segments $OP$ and $OQ$ are perpendicular. In the language of slopes, if $m_1$ is the slope of $OP$ and $m_2$ is the slope of $OQ$, then $m_1 \cdot m_2 = -1$.

Calculating these slopes, we find:

$$m_1 = \frac{t_1 - 0}{t_1^2/2 - 0} = \frac{2}{t_1}, \quad m_2 = \frac{2}{t_2}$$

Multiplying them gives us the elegant condition:

$$t_1 t_2 = -4$$

### The Area and the Algebraic Bridge

The area of a triangle with vertices at the origin and $(x_1, y_1), (x_2, y_2)$ is given by $\frac{1}{2} |x_1 y_2 - x_2 y_1|$. Substituting our parametric coordinates, the expression simplifies to:

$$	ext{Area} = \frac{1}{2} \left| \left( \frac{t_1^2}{2} ight) t_2 - \left( \frac{t_2^2}{2} ight) t_1 ight| = \frac{1}{4} |t_1 t_2 (t_1 - t_2)|$$

Since we know $t_1 t_2 = -4$, the area becomes $|t_1 - t_2|$. Given the area is $3\sqrt{2}$, we have:

$$|t_1 - t_2| = 3\sqrt{2}$$

To find the individual values of $t_1$ and $t_2$, we use the identity $(t_1 + t_2)^2 = (t_1 - t_2)^2 + 4t_1 t_2$. Substituting our known values:

$$(t_1 + t_2)^2 = (3\sqrt{2})^2 + 4(-4) = 18 - 16 = 2$$

This yields two cases: $t_1 + t_2 = \sqrt{2}$ or $t_1 + t_2 = -\sqrt{2}$.

### The Final Reveal

For each case, we form a quadratic equation $t^2 - (t_1+t_2)t + t_1 t_2 = 0$. Solving these, we find the values for $t_1$ and $t_2$.

Since $P$ is in the first quadrant, its $y$-coordinate $t_1$ must be positive. By selecting the positive roots, we arrive at the coordinates:

**$P = (4, 2\sqrt{2})$ and $Q = (1, -\sqrt{2})$** (or vice versa).

Select Answer:

Visualized Solution

Parametric Coordinates on $y^{2} = 2 x$

Circle Diameter and Orthogonality

Slopes of $O P$ and $O Q$

Condition on Parameters $t_{1}$ and $t_{2}$

Area of Triangle $Δ O P Q$

Simplifying the Area Expression

Finding $(t_{1} + t_{2})^{2}$

Solving for $t_{1}$ (Case 1)

Selecting Valid $t_{1}$ for Case 1

Solving for $t_{1}$ (Case 2)

Final Conclusion

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

Analyzing the Setup

The Power of Parametric Elegance

The Orthogonality Insight

The Area and the Algebraic Bridge

The Final Reveal

Similar Questions

Comprehension Passage

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Select Answer:

Visualized Solution

Parametric Coordinates on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=2x

Circle Diameter and Orthogonality

Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OP and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQ

Condition on Parameters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔOPQ