A differential equation representing the family of parabolas with axis parallel to y-axis and whose

Q: Wait, why do we differentiate twice? Is it always necessary?

Great question! In differential equations, the order of the equation corresponds to the number of arbitrary constants you need to eliminate. Here, our general equation $(x - h)^2 = 4a(y - k)$ contains two arbitrary constants, $h$ and $k$. To remove both, we must differentiate twice. If there were only one constant, we would only differentiate once.

Q: Why is the latus rectum $4a$ and not just $a$?

In the standard parabola equation $(x - h)^2 = 4a(y - k)$, the coefficient $4a$ is defined as the length of the latus rectum. It is a standard convention in coordinate geometry. Always be careful to use the full $4a$ value in your equation, not just $a$!

Q: Can we use the vertex form $(y - k)^2 = 4a(x - h)$ instead?

No, you cannot. The problem explicitly states that the axis of the parabola is parallel to the $y$-axis. This means the $x$-term must be squared, leading to the form $(x - h)^2 = 4a(y - k)$. If the axis were parallel to the $x$-axis, then the $y$-term would be squared.

Question

A differential equation representing the family of parabolas with axis parallel to $y$-axis and whose length of latus rectum is the distance of the point $(2, -3)$ from the line $3x + 4y = 5$, is given by :

Options:

(A) $10 \frac{d^2y}{dx^2} = 11$

(B) $11 \frac{d^2x}{dy^2} = 10$

(C) $10 \frac{d^2x}{dy^2} = 11$

(D) $11 \frac{d^2y}{dx^2} = 10$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a family of parabolas. Each one is unique, yet they all share a common trait: their axes are parallel to the $y$-axis.

This means they all open either upwards or downwards, like a fountain frozen in time. There is a hidden constraint—a specific latus rectum length determined by the distance between a point and a line.

### The Bridge: Distance as a Foundation

We start with a point $(2, -3)$ and a line $3x + 4y - 5 = 0$. The problem states that the length of the latus rectum, denoted as $4a$, is exactly the perpendicular distance from this point to the line.

Using the distance formula, $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, we calculate:

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

Simplifying this, we find:

$$4a = \frac{|6 - 12 - 5|}{5} = \frac{|-11|}{5} = \frac{11}{5}$$

This value, $\frac{11}{5}$, is the heartbeat of our parabola family.

### The Canvas: Defining the Family

With $4a = \frac{11}{5}$, we can write the general equation for this family of parabolas:

$$(x - h)^2 = \frac{11}{5}(y - k)$$

Here, $h$ and $k$ are the coordinates of the vertex. Because the vertex can slide anywhere on the plane, $h$ and $k$ are our arbitrary constants.

In the world of differential equations, these constants are like ghosts—they define the shape but must be eliminated to find the underlying law governing the family.

### The Sculptor: Calculus as the Tool

To eliminate these constants, we turn to calculus. First, we differentiate with respect to $x$:

$$\frac{d}{dx}[(x - h)^2] = \frac{d}{dx}\left[\frac{11}{5}(y - k)\right]$$

This yields:

$$2(x - h) = \frac{11}{5} \frac{dy}{dx}$$

Notice how the constant $k$ has vanished. Now, we differentiate once more to remove $h$:

$$\frac{d}{dx}[2(x - h)] = \frac{d}{dx}\left[\frac{11}{5} \frac{dy}{dx}\right]$$

This results in:

$$2 = \frac{11}{5} \frac{d^2y}{dx^2}$$

### The Elegance of the Result

Finally, we rearrange our result to obtain the governing differential equation:

$$10 = 11 \frac{d^2y}{dx^2}$$

Or, expressed in standard form:

**$$11 \frac{d^2y}{dx^2} = 10$$**

This is the differential equation that defines every single parabola in our family. It is a beautiful, simple statement that captures the essence of the geometry we started with.

Animated Solution for Mathematics - Differential Equations: A differential equation representing the family of parabolas with axis parallel to $y$ -axis and whose length of latus rectum is the distance of the point $(2, - 3)$ from the line $3 x + 4 y = 5$ , is given by :

Select Answer:

Visualized Solution

Visualizing the Setup

Distance Formula $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$

Substituting Values

Calculating Latus Rectum $4 a$

Family of Parabolas $(x - h)^{2} = 4 a (y - k)$

Substituting $4 a = \frac{11}{5}$

First Differentiation $\frac{d}{d x}$

Second Differentiation $\frac{d ^{2}}{d x ^{2}}$

Final Differential Equation

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Bridge

The Canvas

The Sculptor

The Elegance of the Result

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Select Answer:

Visualized Solution

Visualizing the Setup

Distance Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=A2+B2​∣Ax1​+By1​+C∣​

Substituting Values

Calculating Latus Rectum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4a

Family of Parabolas .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x−h)2=4a(y−k)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4a=511​

First Differentiation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxd​

Second Differentiation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dx2d2​