The tangents at the point A(1, 3) and B(1, -1) on the parabola y² - 2x - 2y = 1 meet at the point P.

Q: Why can't I just use the slope-point form $y - y_1 = m(x - x_1)$?

You absolutely can, but it is a trap! Finding the slope $m$ requires differentiation or implicit differentiation, which introduces extra steps and potential for calculation errors. The $T=0$ method is a direct, algebraic shortcut that is much safer for competitive exams.

Q: Is there a shortcut for the area of this triangle?

Yes! Since the points $A(1, 3)$ and $B(1, -1)$ have the same x-coordinate, the line segment $AB$ is vertical. Its length is simply the difference in y-coordinates: $|3 - (-1)| = 4$. The height of the triangle is the horizontal distance from $P(-3, 1)$ to the line $x=1$, which is $|1 - (-3)| = 4$. Thus, Area = $\frac{1}{2} \times 4 \times 4 = 8$.

Q: What if I messed up the parabola standardization?

You don't actually need to standardize! The $T=0$ method works on the original equation $y^2 - 2x - 2y - 1 = 0$ just as well. Just remember to replace $y^2$ with $yy_1$, $2x$ with $x+x_1$, and $2y$ with $y+y_1$. The result will be identical.

Question

The tangents at the point $A(1, 3)$ and $B(1, -1)$ on the parabola $y^2 - 2x - 2y = 1$ meet at the point $P$. Then the area (in unit$^2$) of the triangle $PAB$ is :-

Options:

(A) 4

(B) 6

(C) 7

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at the curve defined by $y^2 - 2x - 2y = 1$. It looks like a standard parabola, but it is slightly shifted, hiding its true nature.

To truly understand it, we must first bring it into the light. By rearranging the terms and completing the square, we transform the equation into the elegant standard form:

$$(y - 1)^2 = 2(x + 1)$$

This tells us immediately that the vertex is at $(-1, 1)$ and it opens to the right.

Now, we are given two points, $A(1, 3)$ and $B(1, -1)$. Notice something special? Both points have an $x$-coordinate of $1$.

This means the chord connecting them is a perfectly vertical line. This is a gift from the problem setter—it simplifies our area calculation significantly later on.

### The Power of $T=0$

Now, we need the tangents at these points. You could find the slope by differentiating, but why take the long road when a shortcut exists?

We use the $T=0$ method. This is a fundamental tool in coordinate geometry for any second-degree curve. For our parabola, we replace $y^2$ with $yy_1$, $2y$ with $y+y_1$, and $2x$ with $x+x_1$.

Applying this to our equation $y^2 - 2y = 2x + 1$, the tangent at any point $(x_1, y_1)$ becomes:

$$yy_1 - (y + y_1) = (x + x_1) + 1$$

For point $A(1, 3)$, we substitute $x_1=1$ and $y_1=3$. The equation becomes $3y - (y + 3) = (x + 1) + 1$, which simplifies beautifully to $2y - 3 = x + 2$, or $x - 2y + 5 = 0$. This is our first tangent.

For point $B(1, -1)$, we substitute $x_1=1$ and $y_1=-1$. The equation becomes $-y - (y - 1) = (x + 1) + 1$, which simplifies to $-2y + 1 = x + 2$, or $x + 2y + 1 = 0$. These two lines are the guardians of our triangle.

### The Intersection and the Final Area

Now, we find the point $P$ where these two tangents meet. We have a system of two linear equations: $x - 2y + 5 = 0$ and $x + 2y + 1 = 0$.

Adding them together is a stroke of genius—the $y$ terms vanish instantly! We get $2x + 6 = 0$, which gives $x = -3$.

Substituting this back into the second equation, we find $y = 1$. So, the intersection point $P$ is $(-3, 1)$.

We now have the three vertices of our triangle: $P(-3, 1)$, $A(1, 3)$, and $B(1, -1)$. We could use the determinant formula for the area, but let's use our geometric insight.

The base $AB$ is a vertical segment with length $|3 - (-1)| = 4$. The height of the triangle is the horizontal distance from $P(-3, 1)$ to the line $x=1$, which is $|1 - (-3)| = 4$.

The area is simply:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 4 = 8$$

**The final area is 8 square units.** It is elegant, it is precise, and it is the result of understanding the underlying structure of the conic section.

Animated Solution for Mathematics - Conic Sections: The tangents at the point $A (1, 3)$ and $B (1, - 1)$ on the parabola $y^{2} - 2 x - 2 y = 1$ meet at the point $P$ . Then the area (in unit $^{2}$ ) of the triangle $P A B$ is :-

Select Answer:

Visualized Solution

Visualizing the Parabola and Points

Standardizing the Parabola Equation

The $T = 0$ Method for Tangents

Tangent at Point $A (1, 3)$

Simplifying Tangent at $A$

Tangent at Point $B (1, - 1)$

Simplifying Tangent at $B$

Finding the Intersection Point $P$

Solving for Coordinates of $P$

Setting up the Area of $△ P A B$

Substituting Vertices into Area Formula

Calculating the Final Area

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of $T = 0$

The Intersection and the Final Area

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Visualized Solution

Visualizing the Parabola and Points

Standardizing the Parabola Equation

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0 Method for Tangents

Tangent at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(1,3)

Simplifying Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Tangent at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(1,−1)

Simplifying Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Solving for Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Setting up the Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PAB