Let P(x₀, y₀) be the point on the hyperbola 3x² - 4y² = 36, which is nearest to the line 3x + 2y = 1

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

Minimizing the distance formula $d = |Ax_0 + By_0 + C| / \sqrt{A^2 + B^2}$ subject to the hyperbola constraint is algebraically messy. Using the tangent condition is a much more elegant and standard approach in coordinate geometry.

Q: Why did we get two points? Which one is the nearest?

The hyperbola has two branches. The tangent condition identifies points on both branches where the tangent is parallel to the line. We must calculate the distance of both points to the line to determine which one is truly the nearest.

Q: Is there a faster way than differentiation?

Yes, you could use the condition for tangency $y = mx \pm \sqrt{a^2m^2 - b^2}$. However, differentiation is a universal tool that works for any curve, not just conics, making it a more powerful technique to master.

Question

Let $P(x_0, y_0)$ be the point on the hyperbola $3x^2 - 4y^2 = 36$, which is nearest to the line $3x + 2y = 1$. Then $\sqrt{2}(y_0 - x_0)$ is equal to:

Options:

(A) $-3$

(B) 9

(C) $-9$

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a hyperbola defined by $3x^2 - 4y^2 = 36$. It is a beautiful, sweeping curve, stretching infinitely.

Now, imagine a straight line, $3x + 2y = 1$, cutting across the plane. The problem asks us to find the point on the hyperbola that is closest to this line.

Think of it as a game of approach. Imagine sliding the line parallel to itself, moving it closer and closer to the hyperbola. The very first point on the hyperbola that the line touches is, by definition, the nearest point.

At this precise moment of contact, the line is tangent to the hyperbola. This is our guiding principle: the shortest distance occurs where the tangent to the curve is parallel to the target line.

### The Calculus Bridge

To find this point, we need the slope of the tangent. We have the equation $3x^2 - 4y^2 = 36$.

Let us use the power of calculus to find the slope at any point $(x_0, y_0)$. Differentiating with respect to $x$, we get:

$$6x - 8y \frac{dy}{dx} = 0$$

Rearranging this, we find the slope of the tangent is:

$$\frac{dy}{dx} = \frac{3x}{4y}$$

We know the slope of our target line $3x + 2y = 1$ is $-\frac{3}{2}$. Since the tangent at our point $P(x_0, y_0)$ must be parallel to this line, we set:

$$\frac{3x_0}{4y_0} = -\frac{3}{2}$$

This simplifies beautifully to $6x_0 = -12y_0$, or $x_0 = -2y_0$. This is the algebraic key that unlocks the door.

### The Algebraic Resolution

Now that we have the relationship $x_0 = -2y_0$, we substitute it back into the hyperbola's equation. Replacing $x_0$ with $-2y_0$, we get:

$$3(-2y_0)^2 - 4y_0^2 = 36$$

This simplifies to $3(4y_0^2) - 4y_0^2 = 36$, which is $12y_0^2 - 4y_0^2 = 36$, or $8y_0^2 = 36$. Solving for $y_0^2$, we get:

$$y_0^2 = \frac{36}{8} = \frac{9}{2}$$

Taking the square root, we find $y_0 = \pm \frac{3}{\sqrt{2}}$. This gives us two candidate points: $P_1(-\frac{6}{\sqrt{2}}, \frac{3}{\sqrt{2}})$ and $P_2(\frac{6}{\sqrt{2}}, -\frac{3}{\sqrt{2}})$.

### The Final Verdict

We have two points, but only one is the nearest. We calculate the perpendicular distance from each point to the line $3x + 2y - 1 = 0$.

By testing these, we find that $P_2$ is the point closer to the line. The question asks for the value of $\sqrt{2}(y_0 - x_0)$.

Substituting the coordinates of $P_2$, we calculate:

$$\sqrt{2}\left(-\frac{3}{\sqrt{2}} - \frac{6}{\sqrt{2}}\right) = \sqrt{2}\left(-\frac{9}{\sqrt{2}}\right)$$

The $\sqrt{2}$ terms cancel out, leaving us with exactly **$-9$**. This is the beauty of coordinate geometry: a complex problem reduced to a single, elegant number.

Animated Solution for Mathematics - Conic Sections: Let $P (x_{0}, y_{0})$ be the point on the hyperbola $3 x^{2} - 4 y^{2} = 36$ , which is nearest to the line $3 x + 2 y = 1$ . Then $2 (y_{0} - x_{0})$ is equal to:

Select Answer:

Visualized Solution

Standardizing the Hyperbola

Analyzing the Target Line

The Nearest Point Condition

Differentiation for Slope

Finding the Relation between $x_{0}$ and $y_{0}$

Substitution into Hyperbola

Solving for $y_{0}$

Identifying the Two Points

Distance Comparison

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Calculus Bridge

The Algebraic Resolution

The Final Verdict

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