Question

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

• A. $\sqrt{2}$ units
• B. $3\sqrt{2}$ units
• C. $2\sqrt{2}$ units
• D. $1/\sqrt{2}$ units

Hint:

Whenever dealing with a homogeneous equation of degree 2 (like $ax^2 + 2hxy + by^2 = 0$), try factoring it into two straight lines first. It is drastically faster than using implicit differentiation to find the slope.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a homogenous second-degree equation representing a curve. We must find the equation of the normal line to this curve exactly at the point $(2, 2)$, and then calculate the perpendicular distance from the origin $(0,0)$ to that normal line.

Step 2: Key Formula or Approach:
1. Since the equation is a homogeneous quadratic, it represents a pair of straight lines. We can factor it to find the specific line containing $(2,2)$.
2. The slope of the normal ($m_{\perp}$) is the negative reciprocal of the slope of the line ($m$).
3. Use the point-slope form $y - y_1 = m_{\perp}(x - x_1)$ to find the normal equation.
4. The perpendicular distance from the origin $(0,0)$ to a line $Ax + By + C = 0$ is $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.

Step 3: Detailed Explanation:
Let's factor the curve equation $x^2 + 2xy - 3y^2 = 0$:
$$x^2 + 3xy - xy - 3y^2 = 0$$
$$x(x + 3y) - y(x + 3y) = 0$$
$$(x - y)(x + 3y) = 0$$
The curve consists of two straight lines: $y = x$ and $x = -3y$.
Check which line contains the point $(2, 2)$. It clearly lies on the line $y = x$.
The slope of the line $y = x$ is $m = 1$.
Therefore, the slope of the normal to this line is $m_{\perp} = -1$.
The equation of the normal line passing through $(2, 2)$ with slope $-1$ is:
$$y - 2 = -1(x - 2)$$
$$y - 2 = -x + 2$$
$$x + y - 4 = 0$$
Now, calculate the perpendicular distance from the origin $(0, 0)$ to $x + y - 4 = 0$:
$$d = \frac{|(1)(0) + (1)(0) - 4|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$

Step 4: Final Answer:
The perpendicular distance is $2\sqrt{2}$ units, matching option (c).