Question Analysis and Strategy

We are asked to find the area of the triangle formed by the intersection point of two tangents (P) and their respective points of tangency (A and B) on a given parabola. The problem involves several core concepts from coordinate geometry and conic sections:

1. Standardizing the Parabola Equation: To understand its properties (vertex, axis of symmetry).
2. Equation of Tangent: To find the lines PA and PB.
3. Intersection of Lines: To find point P.
4. Area of a Triangle: To calculate the final answer.

A crucial step will be to observe any symmetry in the points A and B relative to the parabola, as this can significantly simplify finding point P.

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1. Standardizing the Parabola Equation

The first step in analyzing any conic section given in a general form is to convert it into its standard form. This immediately reveals key geometric properties such as the vertex, axis of symmetry, and orientation.

The given equation of the parabola is: $y^{2}-2 x-2 y=1$

Our goal is to transform this into the standard form $(y-k)^2 = 4a(x-h)$ or $(x-h)^2 = 4a(y-k)$. Since the $y^2$ term is present and $x$ is linear, we expect a parabola opening horizontally. We complete the square for the $y$ terms:

Group the $y$ terms: $(y^2 - 2y) - 2x = 1$

To complete the square for $y^2 - 2y$, we add $(\frac{\text{coefficient of } y}{2})^2 = (\frac{-2}{2})^2 = (-1)^2 = 1$ to both sides of the equation. $(y^2 - 2y + 1) - 2x = 1 + 1$

Now, rewrite the perfect square trinomial and rearrange the terms to match the standard form: $(y-1)^2 - 2x = 2$ $(y-1)^2 = 2x + 2$ Factor out the coefficient of $x$ on the right side: $(y-1)^2 = 2(x+1)$

Explanation of Standard Form:

• Comparing this with the standard form $(y-k)^2 = 4a(x-h)$, we can identify:
  • The vertex of the parabola is $(h,k) = (-1, 1)$.
  • The axis of symmetry is the line $y-k=0$, which is $y=1$. This is a horizontal line.
  • The value of $4a=2$, so $a = \frac{1}{2}$. Since $a$ is positive and the squared term is $(y-1)^2$, the parabola opens towards the positive x-direction (to the right).

Tip: Always start by converting to standard form. It's the foundation for understanding the parabola's geometry and can simplify subsequent calculations.

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2. Verifying Points of Tangency and Observing Symmetry

Before proceeding to find the tangents, it's good practice to verify that the given points A and B actually lie on the parabola. This helps confirm the problem setup and ensures our calculations will be valid.

**Verify point A(1,3)$$$$: Substitute $x=1$ and $y=3$ into the standard equation (y-1)^2 = 2(x+1)$:$ (3-1)^2 = 2(1+1) $$ 2^2 = 2(2) $$ 4 = 4 $Since the equation holds true, point$A(1,3)$$ lies on the parabola.

**Verify point B(1,-1)$$$$: Substitute $x=1$ and $y=-1$ into the standard equation (y-1)^2 = 2(x+1)$:$ (-1-1)^2 = 2(1+1) $$ (-2)^2 = 2(2) $$ 4 = 4 $Since the equation holds true, point$B(1,-1)$$ lies on the parabola.

Observing Symmetry: Let's look closely at the coordinates of A and B: $A(1,3)$ and $B(1,-1)$.

• Both points have the same x-coordinate, $x=1$. This means the line segment AB is a vertical line.
• Recall that the axis of symmetry of our parabola is $y=1$.
• Let's check the y-coordinates relative to the axis $y=1$:
  • For A: The y-coordinate is 3. Distance from $y=1$ is $|3-1| = 2$.
  • For B: The y-coordinate is -1. Distance from $y=1$ is $|-1-1| = |-2| = 2$.

Explanation: Since A and B have the same x-coordinate and are equidistant from the horizontal axis of symmetry ($y=1$), they are symmetric with respect to this axis.

Key Concept (Symmetry Property of Parabola): If two points on a parabola are symmetric with respect to its axis of symmetry, then the intersection point of the tangents drawn at these two points will always lie on the axis of symmetry.

This means that the point P (the intersection of tangents at A and B) must lie on the line $y=1$. Therefore, the y-coordinate of P is $y_P = 1$. This observation will significantly simplify finding the coordinates of P.

Tip: Always check for symmetry in geometric problems. It's a powerful tool that can save a lot of calculation time.

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3. Finding the Intersection Point P of the Tangents

We need to find the coordinates of point P where the tangent at A and the tangent at B intersect. We will use the $T=0$ method for finding the tangent equation.

Key Concept (Equation of Tangent, T=0 Method): For a general conic section given by $S(x,y) = 0$, the equation of the tangent at a point $(x_1, y_1)$ on the curve is obtained by making the following substitutions in $S(x,y)$:

• $x^2 \to xx_1$
• $y^2 \to yy_1$
• $x \to \frac{x+x_1}{2}$
• $y \to \frac{y+y_1}{2}$
• $xy \to \frac{xy_1+yx_1}{2}$
• Constants remain unchanged.

The given equation of the parabola is $y^{2}-2 x-2 y-1=0$.

**Equation of Tangent at A(1,3)$$$$ ($x_1=1, y_1=3$): Substitute these values into the $T=0$ transformation: y(3) - 2\left(\frac{x+1}{2}\right) - 2\left(\frac{y+3}{2}\right) - 1 = 0 $$ 3y - (x+1) - (y+3) - 1 = 0 $$ 3y - x - 1 - y - 3 - 1 = 0 $$ -x + 2y - 5 = 0 $Let's rearrange this to a standard linear form:$ x - 2y + 5 = 0 \quad \text{(Equation of Tangent PA)} $$

Using the Symmetry Property to find P (Faster Approach): From our observation in Step 2, we know that point P must lie on the axis of symmetry, $y=1$. Therefore, we already know $y_P=1$. We only need to find $x_P$. Since P lies on the tangent PA (as well as PB), its coordinates must satisfy the equation of tangent PA. Substitute $y=1$ into the equation of tangent PA: $x - 2(1) + 5 = 0$ $x - 2 + 5 = 0$ $x + 3 = 0$ $x = -3$

So, the coordinates of point P are $(-3, 1)$.

Alternative (General Method: Intersection of Two Tangents): If we hadn't noticed the symmetry, we would also find the equation of the tangent at B. **Equation of Tangent at B(1,-1)$$$$ ($x_1=1, y_1=-1$): y(-1) - 2\left(\frac{x+1}{2}\right) - 2\left(\frac{y+(-1)}{2}\right) - 1 = 0 $$ -y - (x+1) - (y-1) - 1 = 0 $$ -y - x - 1 - y + 1 - 1 = 0 $$ -x - 2y - 1 = 0 $Rearranging:$ x + 2y + 1 = 0 \quad \text{(Equation of Tangent PB)} $$

Now, solve the system of equations for tangent PA and tangent PB:

1. $x - 2y + 5 = 0$
2. $x + 2y + 1 = 0$ Add (1) and (2): $(x - 2y + 5) + (x + 2y + 1) = 0$ $2x + 6 = 0$ $2x = -6 \implies x = -3$ Substitute $x=-3$ into Equation (1): $(-3) - 2y + 5 = 0$ $2 - 2y = 0 \implies 2y = 2 \implies y = 1$ Both methods yield the same result: $P(-3, 1)$. The symmetry approach was clearly more efficient.

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4. Calculating the Area of Triangle PAB

Now we have the coordinates of all three vertices of the triangle PAB:

• $P(-3, 1)$
• $A(1, 3)$
• $B(1, -1)$

We can use the formula for the area of a triangle. Given the specific coordinates, where the base AB is a vertical line, the "base times height" formula will be the most straightforward.

Key Concept (Area of a Triangle): If one side of a triangle is parallel to an axis, its area can be calculated as $\frac{1}{2} \times \text{base length} \times \text{height}$.

Calculate the Base AB: The points A and B have the same x-coordinate ($x=1$), so the segment AB is a vertical line. The length of the base AB is the absolute difference in their y-coordinates: $\text{Base AB} = |y_A - y_B| = |3 - (-1)| = |3+1| = 4 \text{ units}$

Calculate the Height: The height of the triangle is the perpendicular distance from the vertex P to the line containing the base AB (which is the line $x=1$). The x-coordinate of P is -3. The x-coordinate of the line AB is 1. $\text{Height} = |x_P - x_{\text{line AB}}| = |-3 - 1| = |-4| = 4 \text{ units}$

Calculate the Area of $\triangle PAB$: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ $\text{Area} = \frac{1}{2} \times 4 \times 4$