Let a line y = mx (m 0) intersect the parabola, y² = x at a point P, other than the origin. Let th

Q: Wait, why is the $x$-coordinate of $Q$ negative? Does that mean the area is negative?

Great observation! The $x$-coordinate of $Q$ is indeed negative, placing it to the left of the origin. However, area is a scalar quantity representing magnitude. When calculating the base of the triangle, we take the absolute value of the coordinate, $|-\frac{1}{m^2}| = \frac{1}{m^2}$, ensuring our area remains positive and physically meaningful.

Q: Is there a faster way to find the $x$-intercept of the tangent?

Absolutely! For any parabola $y^2 = 4ax$, the tangent at $(x_1, y_1)$ always intersects the $x$-axis at $(-x_1, 0)$. This is a powerful property that saves precious time in competitive exams. Once you find $x_1$, you immediately know the $x$-intercept is $-x_1$.

Q: Why can't we just use the formula for the area of a triangle with vertices $(0,0), (x_1, y_1), (x_2, y_2)$?

You absolutely can! The determinant formula $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ works perfectly. However, since one vertex is the origin $(0,0)$ and another lies on the $x$-axis, the base-height method is significantly faster and less prone to algebraic errors.

Question

Let a line $y = mx (m > 0)$ intersect the parabola, $y^2 = x$ at a point $P$, other than the origin. Let the tangent to it at $P$ meet the $x$-axis at the point $Q$. If area $(\Delta OPQ) = 4$ sq. units, then $m$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Intersection

We begin by finding where the line $y = mx$ and the parabola $y^2 = x$ meet. To find the intersection point $P$, we substitute the line equation into the parabola:

$$(mx)^2 = x \implies m^2x^2 - x = 0$$

Do not simply cancel $x$, as this would cause the loss of the origin $O(0,0)$. Instead, factor the expression:

$$x(m^2x - 1) = 0$$

Since $P$ is not the origin, we have $x = \frac{1}{m^2}$. Substituting this back into $y = mx$, we find the $y$-coordinate:

$$y = m \cdot \frac{1}{m^2} = \frac{1}{m}$$

Thus, our point $P$ is located at $(\frac{1}{m^2}, \frac{1}{m})$.

### The Precision Strike of the Tangent

The tangent line captures the instantaneous slope of the curve at point $P$. For a parabola $y^2 = 4ax$, the tangent at $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.

In our case, $y^2 = x$, so $4a = 1$, which implies $a = \frac{1}{4}$. The equation of the tangent becomes:

$$y\left(\frac{1}{m}ight) = \frac{1}{2}\left(x + \frac{1}{m^2}ight)$$

To find where this tangent meets the $x$-axis at point $Q$, we set $y = 0$:

$$0 = \frac{1}{2}\left(x + \frac{1}{m^2}ight) \implies x = -\frac{1}{m^2}$$

Therefore, the coordinates of $Q$ are $(-\frac{1}{m^2}, 0)$.

### The Geometry of Area

We now consider the triangle $\Delta OPQ$ with vertices $O(0,0)$, $P(\frac{1}{m^2}, \frac{1}{m})$, and $Q(-\frac{1}{m^2}, 0)$. The area of a triangle is defined as $\frac{1}{2} \cdot 	ext{base} \cdot 	ext{height}$.

The base $OQ$ lies on the $x$-axis, and its length is the absolute value of the $x$-coordinate of $Q$, which is $\frac{1}{m^2}$. The height is the perpendicular distance from $P$ to the $x$-axis, which is the $y$-coordinate of $P$, $\frac{1}{m}$.

The area calculation is as follows:

$$	ext{Area} = \frac{1}{2} \cdot \left(\frac{1}{m^2}ight) \cdot \left(\frac{1}{m}ight) = \frac{1}{2m^3}$$

### Final Calculation

We are given that the area of the triangle is 4. Setting our expression equal to this value:

$$\frac{1}{2m^3} = 4 \implies \frac{1}{m^3} = 8 \implies m^3 = \frac{1}{8}$$

Taking the cube root of both sides, we arrive at the final result:

**$m = \frac{1}{2}$**

Animated Solution for Mathematics - Conic Sections: Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Finding Intersection Point $P$

Solving for $x$ -coordinate

Finding $y$ -coordinate of $P$

Equation of the Tangent

Applying Tangent Formula

Finding Point $Q$

Identifying Triangle $O P Q$

Area Formula for Triangle

Calculating the Area

Solving for $m$

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Intersection

The Precision Strike of the Tangent

The Geometry of Area

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Curves

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

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P