The distance of the point (6, -2√2) from the common tangent y = mx + c, m 0, of the curves x = 2y²

Q: Wait, why do we use $y = mx + c$ instead of the point-slope form?

Because we do not know the points of tangency yet. The slope-intercept form is the most direct way to relate the tangent to the parabola's parameters. By using $y = mx + c$, we treat the line as a general entity and force it to satisfy the tangency condition for both curves simultaneously.

Q: How do I know if the tangent touches both parabolas at the same time?

By equating the intercepts, we are mathematically forcing the line to be tangent to both curves simultaneously. If the resulting equation for $m$ has a valid solution, then such a common tangent exists. It is a beautiful way of letting the algebra reveal the geometry.

Q: Why did we reject the negative root for $m$?

The problem explicitly states $m > 0$. Always check the constraints given in the question; they are there for a reason! In coordinate geometry, constraints like $m > 0$ often help us select the correct branch of a curve or the correct orientation of a line.

Question

The distance of the point $(6, -2\sqrt{2})$ from the common tangent $y = mx + c, m > 0$, of the curves $x = 2y^2$ and $x = 1 + y^2$ is

Options:

(A) $1/3$

(B) 5

(C) $14/3$

(D) $5\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane, looking at two parabolas. One is $x = 2y^2$, a classic, sleek curve opening to the right, anchored at the origin.

The other is $x = 1 + y^2$, a sibling of the first, but shifted, its vertex resting at $(1, 0)$. They are both reaching out to the right, and our task is to find a single, elegant line—a common tangent—that kisses both of them at the same time.

We are looking for a line $y = mx + c$ that satisfies the tangency condition for both.

### The Tangency Condition: The Key to the Lock

To find this line, we need the magic formula for tangency. For any parabola of the form $y^2 = 4a(x-h)$, a line $y = mx + c$ is tangent if and only if $c = mh + \frac{a}{m}$.

For the first curve, $x = 2y^2$, we rewrite it as $y^2 = \frac{1}{2}x$. Here, the vertex $h$ is $0$, and $4a = \frac{1}{2}$, so $a = \frac{1}{8}$. The condition for tangency becomes:

$$c = m(0) + \frac{1/8}{m} = \frac{1}{8m}$$

For the second curve, $x = 1 + y^2$, we rewrite it as $y^2 = x - 1$. Here, the vertex $h$ is $1$, and $4a = 1$, so $a = \frac{1}{4}$. The condition for tangency becomes:

$$c = m(1) + \frac{1/4}{m} = m + \frac{1}{4m}$$

### The Intersection of Logic

Now, for the line to be a **common tangent**, the $c$ in $y = mx + c$ must be the same for both parabolas. The second equation is $y = m(x-1) + \frac{1}{4m} = mx - m + \frac{1}{4m}$, so the intercept is $-m + \frac{1}{4m}$.

Equating the two expressions for $c$:

$$\frac{1}{8m} = -m + \frac{1}{4m}$$

Rearranging this, we get $m = \frac{1}{4m} - \frac{1}{8m}$, which simplifies to $m = \frac{1}{8m}$. This leads us to $m^2 = \frac{1}{8}$. Since the problem demands $m > 0$, we take the positive root:

$$m = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$$

### The Final Calculation

With $m$ in hand, we find $c = \frac{1}{8(1/2\sqrt{2})} = \frac{2\sqrt{2}}{8} = \frac{1}{2\sqrt{2}}$. Our tangent line is $y = \frac{1}{2\sqrt{2}}x + \frac{1}{2\sqrt{2}}$.

Multiplying by $2\sqrt{2}$ gives $2\sqrt{2}y = x + 1$, or $x - 2\sqrt{2}y + 1 = 0$. Now, we calculate the distance from the point $(6, -2\sqrt{2})$ to this line using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

Substituting our values:

$$d = \frac{|1(6) - 2\sqrt{2}(-2\sqrt{2}) + 1|}{\sqrt{1^2 + (-2\sqrt{2})^2}} = \frac{|6 + 8 + 1|}{\sqrt{1 + 8}} = \frac{15}{3} = 5$$

The distance is exactly **5**. It is a clean, satisfying integer—a reward for our careful navigation through the algebra.

Animated Solution for Mathematics - Conic Sections: The distance of the point $(6, - 22)$ from the common tangent $y = m x + c, m > 0$ , of the curves $x = 2 y^{2}$ and $x = 1 + y^{2}$ is

Select Answer:

Visualized Solution

Visualizing the Curves

Tangent to $x = 2 y^{2}$

Tangent to $x = 1 + y^{2}$

Equating the Intercepts

Calculating Slope $m$

Calculating Intercept $c$

Standard Form of Tangent

Setting up the Distance Formula

Evaluating the Numerator

Evaluating the Denominator

Final Distance Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangency Condition

The Intersection of Logic

The Final Calculation

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