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

Q: Wait, why do we factorize the curve instead of using implicit differentiation?

You absolutely could use implicit differentiation! However, in JEE, time is your greatest currency. Recognizing that the equation is a homogeneous second-degree equation tells us it represents a pair of lines. Factoring it turns a potentially messy derivative problem into a simple geometry problem. Always look for the 'shortcut' that the structure of the equation provides.

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

Think about the definition of a tangent: it is the best linear approximation of a curve at a point. If your 'curve' is actually a straight line, then the line is its own best approximation! Therefore, the tangent at any point on the line is the line itself.

Q: How do I know which line the point $(2, 2)$ belongs to?

Simply substitute the coordinates $(2, 2)$ into both factors. If the result is zero, the point lies on that line. It is a quick verification step that ensures you are working with the correct geometric entity.

Q: Is the normal always perpendicular to the tangent?

Yes, by definition! The normal is the line perpendicular to the tangent at the point of tangency. If the tangent has slope $m$, the normal must have slope $-1/m$. This is a fundamental property that holds true for all smooth curves.

Q: Why is the distance from the origin to the normal $2\sqrt{2}$?

After finding the equation of the normal as $x + y - 4 = 0$, we use the distance formula $d = |Ax_0 + By_0 + C| / \sqrt{A^2 + B^2}$. Plugging in $(0, 0)$ gives $|-4| / \sqrt{1^2 + 1^2}$, which simplifies to $4 / \sqrt{2}$. Rationalizing the denominator gives $2\sqrt{2}$.

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) 2

(B) $2\sqrt{2}$

(C) $4\sqrt{2}$

(D) $\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we encounter a problem that, at first glance, might tempt you to reach for the heavy machinery of calculus.

You see a curve, you see a point, and your instinct screams, 'Differentiate!' But wait—take a breath. In the world of competitive mathematics, the most elegant solutions often come from observation rather than brute force.

Let us look at the equation: 
$$x^2 + 2xy - 3y^2 = 0$$

### The Art of Factorization

This is not just any curve; it is a homogeneous equation of the second degree. It is a secret code, a pair of straight lines passing through the origin, masquerading as a complex curve.

If we split the middle term, $2xy$, into $3xy - xy$, the equation transforms: 
$$x^2 + 3xy - xy - 3y^2 = 0$$

By grouping the terms, we get $x(x + 3y) - y(x + 3y) = 0$, which simplifies beautifully to: 
$$(x + 3y)(x - y) = 0$$

We have unmasked the beast! The 'curve' is actually two distinct lines: 
$L_1: x + 3y = 0$ and $L_2: x - y = 0$.

### Finding Our Place

Now, we must determine which of these lines hosts our point $P(2, 2)$. Testing $L_1$, we find $2 + 3(2) = 8$, which is not zero.

But testing $L_2$, we find $2 - 2 = 0$. The point $P(2, 2)$ lies perfectly on the line $x - y = 0$.

This is a moment of profound simplification. Because the curve at this point is simply a straight line, the tangent to the curve is the line itself. The slope of our tangent, $m_t$, is therefore $1$.

### The Normal and the Final Distance

If the tangent is the line $x - y = 0$, the normal must be the line perpendicular to it passing through $(2, 2)$. Since the slope of the tangent is $1$, the slope of the normal, $m_n$, must be $-1$.

Using the point-slope form, $y - 2 = -1(x - 2)$, we arrive at the equation of the normal: 
$$x + y - 4 = 0$$

Now, we seek the perpendicular distance from the origin $(0, 0)$ to this line. Using the distance formula: 
$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$

We substitute our values: 
$$d = \frac{|1(0) + 1(0) - 4|}{\sqrt{1^2 + 1^2}}$$

This yields: 
$$d = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$

### Reflection

Look at what we have achieved. We didn't need complex derivatives or second-order analysis.

By recognizing the geometric soul of the equation, we turned a daunting problem into a graceful dance of lines. The final answer is **$2\sqrt{2}$**.

Remember, the JEE is not just about solving equations; it is about seeing the patterns beneath them. Keep practicing, keep observing, and keep falling in love with the logic of the universe.

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

Analyze the Curve $x^{2} + 2 x y - 3 y^{2} = 0$

Factorizing the Equation

Identifying the Individual Lines

Testing the Point $(2, 2)$

Slope of the Tangent

Slope of the Normal

Equation of the Normal at $(2, 2)$

Simplifying the Normal Equation

Distance Formula from Origin

Substituting Values

Computing the Final Distance

Summary and Geometric Insight

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Art of Factorization

Finding Our Place

The Normal and the Final Distance

Reflection

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