Let C be the locus of the mirror image of a point on the parabola y² = 4x with respect to the line y

Q: Wait, why do we swap $x$ and $y$ when reflecting across $y = x$?

Think of the line $y = x$ as a diagonal mirror. If you have a point at $(2, 1)$, its mirror image must be at $(1, 2)$. This geometric property holds for every point on the curve. By swapping $x$ and $y$ in the original equation $y^2 = 4x$, we are effectively mapping every point on the original parabola to its reflected counterpart on the new curve $C$.

Q: Can we just use the formula for the tangent to a parabola $y^2 = 4ax$ directly?

Be careful! The formula $y y_1 = 2a(x + x_1)$ applies to the parabola $y^2 = 4ax$. If you use it on the original parabola $y^2 = 4x$ at point $(2, 1)$, you will get the tangent to the *original* parabola, not the reflected one. You must first find the equation of the reflected curve $C$ and then find the tangent to that specific curve.

Q: Is there a faster way to find the tangent without differentiating?

Yes, you can use the $T=0$ method for conic sections. For a curve $x^2 = 4y$, the tangent at $(x_1, y_1)$ is given by $x x_1 = 2(y + y_1)$. Substituting $(2, 1)$ into this formula gives $2x = 2(y + 1)$, which simplifies to $x = y + 1$, or $x - y = 1$. Both calculus and the $T=0$ method lead to the same beautiful result!

Question

Let $C$ be the locus of the mirror image of a point on the parabola $y^2 = 4x$ with respect to the line $y = x$. Then the equation of tangent to $C$ at $P(2, 1)$ is :

Options:

(A) x - y = 1

(B) 2x + y = 5

(C) x + 3y = 5

(D) x + 2y = 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a mirror, but not just any mirror. You are standing before the line $y = x$, a perfect diagonal divider of the coordinate plane.

The original parabola, $y^2 = 4x$, is a classic, opening gracefully to the right. When we reflect it across the line $y = x$, we are performing a fundamental coordinate transformation.

The golden rule of reflection across $y = x$ is the simple swap: $(x, y) \rightarrow (y, x)$. By applying this to our original equation $y^2 = 4x$, we replace $y$ with $x$ and $x$ with $y$.

The result is a new, elegant curve $C$ defined by:

$$x^2 = 4y$$

This is an upward-opening parabola, a perfect mirror image of our original.

### The Calculus Bridge: Finding the Slope

Now that we have identified our curve $C$ as $x^2 = 4y$, we turn our attention to the point $P(2, 1)$. First, let us verify that $P$ indeed lies on our new curve.

Substituting $x = 2$ into $x^2 = 4y$, we get $4 = 4y$, which means $y = 1$. It fits perfectly!

Now, we need the tangent line at this point. In the language of calculus, the slope of the tangent is the derivative $\frac{dy}{dx}$. We take our equation $x^2 = 4y$ and differentiate both sides with respect to $x$:

$$2x = 4 \frac{dy}{dx}$$

Solving for the slope, we find:

$$\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$$

This expression, $\frac{x}{2}$, is our slope generator. It tells us the slope of the tangent at any point on the curve $C$.

### The Final Construction

We are almost there. We need the slope at our specific point $P(2, 1)$. By substituting $x = 2$ into our slope formula $\frac{x}{2}$, we get:

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

The slope of our tangent is $1$. With the slope $m = 1$ and the point $(2, 1)$ in hand, we invoke the point-slope form of a line: $y - y_1 = m(x - x_1)$.

Plugging in our values, we get $y - 1 = 1(x - 2)$. Simplifying this, we arrive at $y - 1 = x - 2$, which rearranges beautifully into:

**$x - y = 1$**

This is the equation of the tangent line to curve $C$ at $P(2, 1)$. You have successfully navigated the reflection, the differentiation, and the final linear construction.

Animated Solution for Mathematics - Conic Sections: Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

The Reflection Property

Transforming the Equation

Identifying Curve $C$

Locating Point $P (2, 1)$

Finding the Slope

Differentiating the Equation

Solving for $\frac{d y}{d x}$

Calculating Slope at $P$

Point-Slope Form Setup

Expanding the Equation

Final Simplification

The Way Forward

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Calculus Bridge

The Final Construction

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Select Answer:

Visualized Solution

Visualizing the Setup

The Reflection Property

Transforming the Equation

Identifying Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Locating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2,1)

Finding the Slope

Differentiating the Equation

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

Calculating Slope at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P