A line passing through the point A(-2, 0), touches the parabola P : y² = x - 2 at the point B in the

Q: Wait, why do we integrate with respect to $y$ instead of $x$?

Great question! If you integrate with respect to $x$, you would have to split the region into two parts because the lower boundary changes from the $x$-axis to the parabola at $x=2$. Integrating with respect to $y$ allows you to treat the entire region as a single integral of $(x_{right} - x_{left}) dy$, which is much cleaner and less prone to calculation errors.

Q: Why did we choose $m = \frac{1}{4}$ and not $-\frac{1}{4}$?

The problem explicitly states that the point of contact $B$ is in the first quadrant. A negative slope would result in a tangent line that touches the parabola in the fourth quadrant (where $y < 0$). Since we need the first quadrant, we must select the positive slope.

Q: Is there a shortcut for the area of a parabolic segment?

Yes, there is! The area of a parabolic segment cut by a chord is $\frac{4}{3}$ times the area of the triangle formed by the chord and the tangents at the endpoints. However, for JEE Advanced, mastering the integration method is crucial because it works for any curve, not just parabolas. Stick to the integration method for now to build your conceptual foundation!

Question

A line passing through the point $A(-2, 0)$, touches the parabola $P : y^2 = x - 2$ at the point $B$ in the first quadrant. The area, of the region bounded by the line $AB$, parabola $P$ and the $x$-axis, is :-

Options:

(A) $\frac{7}{3}$

(B) $\frac{8}{3}$

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the area bounded by the tangent line drawn from point $A(-2, 0)$ to the parabola $P$ defined by $y^2 = x - 2$, the parabola itself, and the $x$-axis. The parabola $P$ has its vertex at $(2, 0)$ and opens to the right.

### The Algebraic Dance

To find the tangent line, we assume its slope is $m$. Since the line passes through $A(-2, 0)$, its equation is:

$$y = m(x + 2)$$

We substitute this into the parabola's equation $y^2 = x - 2$ to find the intersection:

$$(m(x + 2))^2 = x - 2$$

Expanding this yields:

$$m^2(x^2 + 4x + 4) = x - 2 \Rightarrow m^2x^2 + (4m^2 - 1)x + (4m^2 + 2) = 0$$

Since the line is tangent, the discriminant $D$ of this quadratic must be zero:

$$D = (4m^2 - 1)^2 - 4(m^2)(4m^2 + 2) = 0$$

Expanding the discriminant, the $16m^4$ terms cancel, leaving:

$$1 - 8m^2 - 8m^2 = 0 \Rightarrow 1 - 16m^2 = 0 \Rightarrow m^2 = \frac{1}{16}$$

Choosing the positive slope for the first quadrant contact, we have $m = \frac{1}{4}$.

### Finding the Point of Contact

With $m = \frac{1}{4}$, the line equation becomes $y = \frac{1}{4}(x + 2)$, which rearranges to $x = 4y - 2$. Substituting this into the parabola $y^2 = x - 2$:

$$y^2 = (4y - 2) - 2 \Rightarrow y^2 - 4y + 4 = 0$$

This is a perfect square:

$$(y - 2)^2 = 0 \Rightarrow y = 2$$

Substituting $y=2$ back into the line equation, we find $x = 6$. Thus, the point of contact $B$ is $(6, 2)$.

### The Elegance of Integration

To find the area, we integrate with respect to $y$ from $y = 0$ to $y = 2$. The area is defined by the difference between the parabola $x = y^2 + 2$ and the line $x = 4y - 2$:

$$	ext{Area} = \int_{0}^{2} ((y^2 + 2) - (4y - 2)) \, dy$$

Simplifying the integrand:

$$	ext{Area} = \int_{0}^{2} (y^2 - 4y + 4) \, dy = \int_{0}^{2} (y - 2)^2 \, dy$$

Evaluating the definite integral:

$$\left[ \frac{(y - 2)^3}{3} ight]_{0}^{2} = 0 - \left( \frac{(0 - 2)^3}{3} ight) = - \left( \frac{-8}{3} ight) = \frac{8}{3}$$

The final area is **$\frac{8}{3}$ square units**.

Animated Solution for Mathematics - Definite Integration: A line passing through the point $A (- 2, 0)$ , touches the parabola $P : y^{2} = x - 2$ at the point $B$ in the first quadrant. The area, of the region bounded by the line $A B$ , parabola $P$ and the $x$ -axis, is :-

Select Answer:

Visualized Solution

Visualize the Parabola and Point $A$

Equation of the Tangent Line

Substitute Line into Parabola

Form the Quadratic Equation

Condition for Tangency ( $D = 0$ )

Solve for the Slope $m$

Determine Point of Contact $B$

Identify the Shaded Region

Set up the Area Integral

Simplify the Integrand

Execute the Integration

Substitute Limits and Calculate

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Algebraic Dance

Finding the Point of Contact

The Elegance of Integration

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Select Answer:

Visualized Solution

Visualize the Parabola and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Equation of the Tangent Line

Substitute Line into Parabola

Form the Quadratic Equation

Condition for Tangency ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=0)

Solve for the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Determine Point of Contact .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B