Question

The equation of normal to the curve $y=\frac{2}{x^{2}}$ at the point on the curve where $x=1$, is

• A. $4y-x-7=0$
• B. $y-4x+2=0$
• C. $4y+x-9=0$
• D. $y-x-1=0$
• E. $4y+x+7=0$

Hint:

Geometry Tip: You can verify your line equation by plugging the original point $(1, 2)$ back into it: $4(2) - (1) - 7 = 8 - 1 - 7 = 0$.

Answer 1

The Correct Option is A

Concept:
The normal line is perfectly perpendicular to the tangent line at a given point. Its slope ($m_n$) is the negative reciprocal of the tangent's slope ($m_t$). Once the normal slope and the point coordinates are found, use the point-slope form $y - y_1 = m_n(x - x_1)$ to construct the final equation.

Step 1: Find the y-coordinate of the point.
Substitute $x = 1$ into the original curve equation: $$y = \frac{2}{(1)^2} = 2$$ The point of intersection is $(1, 2)$.

Step 2: Find the derivative (slope of the tangent).
Rewrite the function as $y = 2x^{-2}$ and apply the power rule: $$\frac{dy}{dx} = 2(-2x^{-3}) = -\frac{4}{x^3}$$ Evaluate at $x = 1$ to find the tangent slope $m_t$: $$m_t = -\frac{4}{(1)^3} = -4$$

Step 3: Calculate the slope of the normal line.
The normal slope is the negative reciprocal of the tangent slope: $$m_n = -\frac{1}{m_t}$$ $$m_n = -\left(\frac{1}{-4}\right) = \frac{1}{4}$$

Step 4: Set up the point-slope equation.
Use the point $(1, 2)$ and the normal slope $m_n = \frac{1}{4}$: $$y - y_1 = m_n(x - x_1)$$ $$y - 2 = \frac{1}{4}(x - 1)$$

Step 5: Simplify into standard form.
Multiply the entire equation by 4 to remove the fraction: $$4(y - 2) = 1(x - 1)$$ $$4y - 8 = x - 1$$ Rearrange to group all terms on the left side: $$4y - x - 7 = 0$$ Hence the correct answer is (A) $4y-x-7=0$.