Find a point on the curve x² + 2y² = 6 whose distance from the line x + y = 7 is minimum.

Q: Why do we set the derivative equal to the slope of the line?

Great question! The shortest distance from a point on a curve to a line is measured along the perpendicular. If you imagine a line parallel to our given line $x + y = 7$ moving towards the ellipse, the first point of contact is where the distance is minimized. At this point of contact, the tangent to the ellipse must be parallel to the line. Since parallel lines share the same slope, we set the derivative $\frac{dy}{dx}$ equal to the slope of the line, which is $-1$.

Q: Why did we get two points, $(2, 1)$ and $(-2, -1)$? Which one is the minimum?

The derivative condition $\frac{dy}{dx} = -1$ identifies points where the tangent is parallel to the line. On an ellipse, there are two such points: one on the side closest to the line and one on the side farthest from the line. We calculated the perpendicular distance for both points to the line $x + y - 7 = 0$ and found that $(2, 1)$ gives the smaller distance, $2\sqrt{2}$, while $(-2, -1)$ gives the larger distance, $5\sqrt{2}$. Thus, $(2, 1)$ is our winner!

Q: Can we use the distance formula directly without derivatives?

You could, but it would be algebraically messy! You would have to express the distance from a general point $(x, y)$ on the ellipse to the line as a function of one variable, say $y$, using the constraint $x^2 + 2y^2 = 6$. This would involve square roots and potentially complex optimization. Using the parallel tangent condition is a much more elegant and standard approach in coordinate geometry for finding extreme distances.

Question

Find a point on the curve $x^2 + 2y^2 = 6$ whose distance from the line $x + y = 7$ is minimum.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful landscape of coordinate geometry. Today, we are not just solving an equation; we are exploring the relationship between a graceful curve and a straight path.

Imagine an elliptical track defined by the equation $x^2 + 2y^2 = 6$. Nearby, there is a straight highway, represented by the line $x + y = 7$.

Our mission is to find the exact spot on this elliptical track that is closest to the highway. This is a classic problem that bridges the gap between pure algebra and geometric intuition.

### The Geometric Intuition

Before we dive into the calculus, let's pause and visualize. If you were standing on this elliptical track, looking for the point closest to the highway, how would you identify it?

Imagine a line parallel to our highway, $x + y = 7$, sliding across the plane towards the origin. As it moves, it will eventually graze the ellipse at a single, perfect point. That point of contact is the closest point!

Because the sliding line is parallel to our highway, the tangent to the ellipse at that point must also be parallel to the highway. This is our golden key: the slope of the tangent at the closest point must equal the slope of the line $x + y = 7$. Since the line can be rewritten as $y = -x + 7$, its slope is clearly $m = -1$.

### The Calculus Engine

Now, let's bring in the power of calculus to find this point. We need to find the slope of the tangent at any point $(x, y)$ on the ellipse $x^2 + 2y^2 = 6$. We do this by differentiating with respect to $x$:

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(2y^2) = \frac{d}{dx}(6)$$

Applying the power rule and the chain rule, we get:

$$2x + 4y \frac{dy}{dx} = 0$$

This gives us the general expression for the slope of the tangent: $\frac{dy}{dx} = -\frac{2x}{4y} = -\frac{x}{2y}$. We know that at our target point, this slope must be $-1$.

So, we set our derivative equal to $-1$:

$$-\frac{x}{2y} = -1 \implies x = 2y$$

This simple, elegant relation, $x = 2y$, is the geometric condition that our point must satisfy. It tells us that for any point on the ellipse to be a candidate for the closest distance, its $x$-coordinate must be exactly twice its $y$-coordinate.

### The Intersection

Now, we bring this condition back to the ellipse itself. We substitute $x = 2y$ into the original equation $x^2 + 2y^2 = 6$:

$$(2y)^2 + 2y^2 = 6$$

$$4y^2 + 2y^2 = 6$$

$$6y^2 = 6 \implies y^2 = 1$$

This gives us two possible values for $y$: $y = 1$ and $y = -1$. If $y = 1$, then $x = 2(1) = 2$, giving us the point $(2, 1)$. If $y = -1$, then $x = 2(-1) = -2$, giving us the point $(-2, -1)$.

### The Final Verdict

We have two candidate points, but which one is the closest? We use the perpendicular distance formula from a point $(x_1, y_1)$ to the line $ax + by + c = 0$, which is defined as:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

For our line $x + y - 7 = 0$, we calculate the distances for both candidates:

For $(2, 1)$: 
$$d_1 = \frac{|2 + 1 - 7|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{2}} = 2\sqrt{2}$$

For $(-2, -1)$: 
$$d_2 = \frac{|-2 - 1 - 7|}{\sqrt{1^2 + 1^2}} = \frac{|-10|}{\sqrt{2}} = 5\sqrt{2}$$

Comparing the two, $2\sqrt{2}$ is clearly smaller than $5\sqrt{2}$. Therefore, the **closest point on the ellipse to the line is $(2, 1)$**.

Animated Solution for Mathematics - Differentiation: Find a point on the curve $x^{2} + 2 y^{2} = 6$ whose distance from the line $x + y = 7$ is minimum.

Visualized Solution

Visualizing the Curve and Line

The Parallel Tangent Condition

Differentiating $x^{2} + 2 y^{2} = 6$

Equating Slopes: $\frac{d y}{d x} = - 1$

Substituting $x = 2 y$

Solving for $(x, y)$

Distance Check Setup

Evaluating $d_{1}$ and $d_{2}$

Conclusion and Takeaway

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Geometric Intuition

The Calculus Engine

The Intersection

The Final Verdict

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find a point on the curve x2+2y2=6 whose distance from the line x+y=7 is minimum.

Visualized Solution

Visualizing the Curve and Line

The Parallel Tangent Condition

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+2y2=6

Equating Slopes: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​=−1

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2y

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