If the tangent at a point P on the parabola y² = 3x is parallel to the line x + 2y = 1 and the tange

Q: Wait, why do we reject the negative root for the parabola tangent? Or do we?

Great observation! Actually, we don't reject it arbitrarily. We solve the equation $\frac{3}{2y} = -\frac{1}{2}$, which leads to a unique value for $y$. If the math had given us a quadratic in $y$, we would have had to check both points. Always let the algebra dictate the geometry!

Q: Why can't we just use the distance formula for the area?

You absolutely can! You could find the base $QR$ and the height from $P$ to the line $QR$. However, the coordinate geometry formula (the determinant method) is often faster and less prone to calculation errors when you have the coordinates of all three vertices. It's a powerful tool in your arsenal.

Q: Why are there two tangents for the ellipse but only one for the parabola?

That's a fundamental property of these curves! A line with a given slope can be tangent to an ellipse at two opposite points (because the ellipse is a closed, symmetric curve). For a parabola, a line with a given slope is tangent at exactly one point because the parabola is open and extends to infinity.

Question

If the tangent at a point P on the parabola $y^2 = 3x$ is parallel to the line $x + 2y = 1$ and the tangents at the points Q and R on the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ are perpendicular to the line $x - y = 2$, then the area of the triangle PQR is:

Options:

(A) $\frac{9}{\sqrt{5}}$

(B) $5\sqrt{3}$

(C) $\frac{3}{2}\sqrt{5}$

(D) $3\sqrt{5}$

Sigma Percentile · Accepted Answer

### The Dance of Curves and Tangents

Welcome, fellow explorer of the mathematical universe! Today, we are not just solving a problem; we are choreographing a dance between a parabola and an ellipse.

Imagine standing on a vast coordinate plane. On one side, you see the graceful curve of the parabola $y^2 = 3x$. On the other, the elegant, closed loop of the ellipse:

$$ \frac{x^2}{4} + \frac{y^2}{1} = 1 $$

Our mission is to find the area of a triangle formed by three specific points: $P$ on the parabola, and $Q$ and $R$ on the ellipse. Let's break this down, step by step.

### Phase 1: The Parabola and Point $P$

We begin with the parabola $y^2 = 3x$. We are told that the tangent at point $P$ is parallel to the line $x + 2y = 1$.

Now, pause and think: what does 'parallel' mean in the language of slopes? It means the slopes are identical. The line $x + 2y = 1$ can be rewritten as $y = -\frac{1}{2}x + \frac{1}{2}$, so its slope is $m = -\frac{1}{2}$.

Thus, the tangent at $P$ must also have a slope of $m = -\frac{1}{2}$. To find the coordinates of $P$, we differentiate $y^2 = 3x$ with respect to $x$:

$$ 2y \frac{dy}{dx} = 3 \Rightarrow \frac{dy}{dx} = \frac{3}{2y} $$

Equating this to our required slope, we get:

$$ \frac{3}{2y_P} = -\frac{1}{2} \Rightarrow y_P = -3 $$

Substituting $y_P = -3$ back into the parabola's equation, $(-3)^2 = 3x_P$, we find $x_P = 3$. Our first vertex, $P$, is $(3, -3)$.

### Phase 2: The Ellipse and Points $Q$ and $R$

Now, shift your gaze to the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$. We are looking for points $Q$ and $R$ where the tangents are perpendicular to the line $x - y = 2$.

The slope of the line $x - y = 2$ is $1$. Since the tangents are perpendicular, their slope must be the negative reciprocal, $m = -1$.

We use the standard condition for a line $y = mx + c$ to be tangent to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which is $c = \pm \sqrt{a^2m^2 + b^2}$. Here, $a^2 = 4$, $b^2 = 1$, and $m = -1$.

Substituting these, we get:

$$ c = \pm \sqrt{4(-1)^2 + 1} = \pm \sqrt{5} $$

So, our two tangent lines are $y = -x + \sqrt{5}$ and $y = -x - \sqrt{5}$. To find the points of contact $(x, y)$, we use the formula:

$$ x = \mp \frac{a^2m}{c}, \quad y = \pm \frac{b^2}{c} $$

For $c = \sqrt{5}$, we get $Q = (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$. For $c = -\sqrt{5}$, we get $R = (-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}})$.

Notice the symmetry! $R$ is simply $-Q$, which will make our area calculation much smoother.

### Phase 3: The Area of Triangle $PQR$

We have our vertices: $P(3, -3)$, $Q(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$, and $R(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}})$. The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:

$$ 	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $$

Because $R = -Q$, the expression simplifies significantly to the form $|x_P y_Q - x_Q y_P|$. Substituting our values:

$$ 	ext{Area} = \left| 3\left(\frac{1}{\sqrt{5}}ight) - \left(\frac{4}{\sqrt{5}}ight)(-3) ight| = \left| \frac{3}{\sqrt{5}} + \frac{12}{\sqrt{5}} ight| = \frac{15}{\sqrt{5}} $$

Rationalizing this, we get:

$$ \frac{15\sqrt{5}}{5} = 3\sqrt{5} $$

And there it is! The geometry and algebra have converged perfectly. The **area of triangle $PQR$ is $3\sqrt{5}$**.

Select Answer:

Visualized Solution

Visualizing the Parabola and Reference Line

Slope of the Tangent at $P$

Differentiating to Find Slope

Calculating Coordinates of Point $P$

Analyzing the Ellipse and Second Reference Line

Slope of Tangents at $Q$ and $R$

Equation of Tangents to the Ellipse

Finding Points $Q$ and $R$

Setting up the Area of Triangle $P QR$

Applying the Area Formula

Simplifying the Area Expression

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Dance of Curves and Tangents

Phase 1

Phase 2

Phase 3

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