The length of the perpendicular from the origin on the normal to the curve, x² + 2xy - 3y² = 0 at th

Q: Wait, why do we factorize the equation? Can't we just use implicit differentiation to find the slope?

You absolutely could use implicit differentiation, and it would yield the same result! However, recognizing the homogeneous form is a powerful shortcut. It reveals the underlying geometry immediately, showing us that the curve is just a pair of lines. In JEE Advanced, spotting these geometric shortcuts saves precious time and reduces the chance of calculation errors.

Q: Why is the tangent to the curve just the line itself?

Think about the definition of a tangent. It is the line that best approximates the curve at a point. If the curve is already a straight line, the best approximation is the line itself! So, at any point on the line $y = x$, the tangent is simply $y = x$.

Q: Is the distance formula $d = \frac{|C|}{\sqrt{A^2 + B^2}}$ always valid for the origin?

Yes, it is! The general formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. When the point is the origin $(0, 0)$, the $Ax_1$ and $By_1$ terms vanish, leaving you with the elegant $\frac{|C|}{\sqrt{A^2 + B^2}}$.

Question

The length of the perpendicular from the origin on the normal to the curve, $x^2 + 2xy - 3y^2 = 0$ at the point $(2, 2)$ is

Options:

(A) $4\sqrt{2}$

(B) 2

(C) $2\sqrt{2}$

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the equation $x^2 + 2xy - 3y^2 = 0$. At first glance, it might look like a daunting quadratic form, but for the trained eye of a JEE aspirant, this is a beautiful, symmetric structure.

This is a homogeneous equation of degree 2. In the world of coordinate geometry, such equations are special; they represent a pair of straight lines that intersect at the origin.

This transforms a potentially messy calculus problem into a clean, geometric one. We do not need to dive into complex derivatives if we can simply unmask the lines hidden within.

### Unmasking the Lines

To find these lines, we perform a simple algebraic maneuver: factorization. We look at the middle term, $2xy$, and split it to group the terms effectively.

We rewrite the equation as:
$$x^2 + 3xy - xy - 3y^2 = 0$$

Now, we group them: $x(x + 3y) - y(x + 3y) = 0$. This leads us to the factored form:
$$(x - y)(x + 3y) = 0$$

Just like that, the mystery is solved. Our curve is actually two distinct lines: $y = x$ and $x + 3y = 0$.

### The Tangent and the Normal

Now, let us focus on the point $P(2, 2)$. We need to find the normal at this point.

First, we must determine which of our two lines contains $P$. Substituting $x = 2$ and $y = 2$ into $x - y = 0$, we get $2 - 2 = 0$. It fits perfectly; our point $P$ lies on the line $y = x$.

Here is the conceptual leap: since the curve at this point is just a straight line, the tangent to the curve is the line itself. The slope of the tangent, $m_T$, is therefore the slope of the line $y = x$, which is $1$.

The normal, by definition, is perpendicular to the tangent. Its slope, $m_N$, is the negative reciprocal of the tangent's slope:
$$m_N = -\frac{1}{m_T} = -1$$

With the point $(2, 2)$ and the slope $m_N = -1$, we use the point-slope form: $y - 2 = -1(x - 2)$. Simplifying this, we get $y - 2 = -x + 2$, which rearranges to the elegant normal equation:
$$x + y - 4 = 0$$

### The Final Distance

We have arrived at the final stage of our journey. We need the perpendicular distance from the origin $(0, 0)$ to our normal line $x + y - 4 = 0$.

We use the standard distance formula:
$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substituting our values, where $A = 1$, $B = 1$, $C = -4$, and $(x_1, y_1) = (0, 0)$, we get:
$$d = \frac{|1(0) + 1(0) - 4|}{\sqrt{1^2 + 1^2}}$$

This simplifies to $d = \frac{|-4|}{\sqrt{2}}$, which is $\frac{4}{\sqrt{2}}$. Rationalizing the denominator, we multiply the numerator and denominator by $\sqrt{2}$ to get:
$$d = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$$

And there it is! A beautiful, precise result. The perpendicular distance from the origin to the normal is **$2\sqrt{2}$**.

Animated Solution for Mathematics - Straight Lines: The length of the perpendicular from the origin on the normal to the curve, $x^{2} + 2 x y - 3 y^{2} = 0$ at the point $(2, 2)$ is

Select Answer:

Visualized Solution

The Given Curve

Factorizing the Equation

Grouping Terms

The Two Straight Lines

Locating the Point $P (2, 2)$

Slope of the Tangent

Slope of the Normal

Equation of the Normal

Simplifying the Normal Equation

Distance from Origin

Substituting the Values

Calculating the Distance

Final Answer

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

Unmasking the Lines

The Tangent and the Normal

The Final Distance

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the perpendicular from the origin on the normal to the curve, x2+2xy−3y2=0 at the point (2,2) is