Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to

Q: Wait, why do we reject the negative root here? Or rather, why did we reject $k=12$?

That is a fantastic question! Remember, the directrix is a line that, by definition, does not intersect the parabola. If you plug the focus $(4, 4)$ into the equation $x + 2y = 12$, you get $4 + 2(4) = 12$, which means the line passes through the focus. A directrix can never pass through the focus, so we must reject $k=12$.

Q: How do I know the directrix is parallel to the mirror line?

Look at the slopes! The axis of the parabola passes through $(3, 2)$ and $(4, 4)$, so its slope is $m = \frac{4-2}{4-3} = 2$. The directrix is perpendicular to the axis, so its slope is $-1/m = -1/2$. The mirror line $x + 2y = 6$ can be rewritten as $y = -1/2x + 3$, which also has a slope of $-1/2$. Since they share the same slope, they are parallel.

Q: Why can't we just reflect the vertex and focus and then find the new parabola?

You absolutely could! That is a perfectly valid method. However, reflecting the directrix is much faster. In competitive exams like JEE, time is your most valuable asset. Using the property of parallel lines and arithmetic progressions saves you from complex coordinate geometry calculations.

Question

Let $P_1$ be a parabola with vertex $(3, 2)$ and focus $(4, 4)$ and $P_2$ be its mirror image with respect to the line $x + 2y = 6$. Then the directrix of $P_2$ is $x + 2y = ____$.

Sigma Percentile · Accepted Answer

### The Geometry of Reflection: A Parabolic Journey

Imagine you are standing on a coordinate plane. You have a parabola, $P_1$, defined by its vertex at $V(3, 2)$ and its focus at $S(4, 4)$.

We are tasked with finding the directrix of its mirror image, $P_2$, reflected across the line $x + 2y = 6$. This is a journey of geometric elegance.

### Phase 1: The Anatomy of $P_1$

Every parabola has an axis of symmetry passing through both the vertex $V(3, 2)$ and the focus $S(4, 4)$. To find its slope, we calculate the change in $y$ over the change in $x$:

$$m_{axis} = \frac{4 - 2}{4 - 3} = 2$$

The directrix is always perpendicular to the axis of symmetry. Since the axis has a slope of $2$, the directrix must have a slope of $-\frac{1}{2}$.

This implies the equation of our directrix takes the form $x + 2y = k$. Our goal is to determine the constant $k$.

### Phase 2: The Distance Logic

The distance from the vertex to the focus, denoted as $a$, is equal to the distance from the vertex to the directrix. We calculate $a$ using the distance formula between $V(3, 2)$ and $S(4, 4)$:

$$a = \sqrt{(4-3)^2 + (4-2)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$$

Next, we use the perpendicular distance formula from the vertex $(3, 2)$ to the line $x + 2y - k = 0$:

$$\frac{|3 + 2(2) - k|}{\sqrt{1^2 + 2^2}} = \sqrt{5}$$

Simplifying this expression, we obtain:

$$\frac{|7 - k|}{\sqrt{5}} = \sqrt{5} \Rightarrow |7 - k| = 5$$

This yields two possibilities: $7 - k = 5$ (so $k = 2$) or $7 - k = -5$ (so $k = 12$). We reject $k = 12$ because it would force the directrix to pass through the focus. Thus, our original directrix $D_1$ is $x + 2y = 2$.

### Phase 3: The Reflection

We reflect $P_1$ across the line $L: x + 2y = 6$. Note that $L$ is parallel to $D_1$.

When reflecting a line across another line parallel to it, the resulting line remains parallel. The mirror line $L$ acts as the geometric midpoint between the original directrix $D_1$ and the new directrix $D_2$.

Because $L$ is the midpoint, the constants in their equations form an arithmetic progression. If the constant of $D_2$ is $k'$, then the average of $2$ and $k'$ must be $6$:

$$\frac{2 + k'}{2} = 6$$

Solving for $k'$, we find $2 + k' = 12$, which results in $k' = 10$.

### Final Calculation

The directrix of the reflected parabola $P_2$ is defined by the equation:

**$x + 2y = 10$**

Animated Solution for Mathematics - Conic Sections: Let $P_{1}$ be a parabola with vertex $(3, 2)$ and focus $(4, 4)$ and $P_{2}$ be its mirror image with respect to the line $x + 2 y = 6$ . Then the directrix of $P_{2}$ is $x + 2y = ____$ .

Enter Numerical Value:

Visualized Solution

Visualizing Parabola $P_{1}$

Slope of the Axis

Slope of the Directrix

Distance from Vertex to Focus ( $a$ )

Vertex to Directrix Distance

Applying the Distance Formula

Solving for $k$

Selecting the Correct Directrix

The Mirror Line $L$

Reflecting the Directrix

Equidistant Parallel Lines

Calculating $k^{'}$

Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Geometry of Reflection

Phase 1

Phase 2

Phase 3

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​