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 Concept:
First, find the slope of the tangent at $(2, 2)$, then the slope of the normal. Use the point-slope form to find the equation of the normal.

Step 2: Formula Application:
Differentiating $x^2 + 2xy - 3y^2 = 0$: $2x + 2(x y' + y) - 6y y' = 0$. The length of perpendicular from $(0, 0)$ to $ax + by + c = 0$ is $d = \frac{|c|}{\sqrt{a^2+b^2}}$.

Step 3: Explanation:
At $(2, 2)$: $4 + 2(2y' + 2) - 12y' = 0 \implies 4 + 4y' + 4 - 12y' = 0 \implies 8 = 8y' \implies y' = 1$. Slope of tangent $= 1$, so slope of normal $m_n = -1$. Equation of normal: $y - 2 = -1(x - 2) \implies x + y - 4 = 0$. Perpendicular distance from $(0, 0) = \frac{|-4|}{\sqrt{1^2+1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

Step 4: Final Answer:
The length is $2\sqrt{2}$ units.