Question

The normal to the curve y=√x at the point (25, 5) intersects the y-axis at

• A. (0,245)
• B. (0,255)
• C. (255,0)
• D. (245,0)
• E. (0,100)

Answer 1

The Correct Option is B

We are given the curve $y = \sqrt{x}$ and the point of tangency $(25, 5)$. The goal is to find the equation of the normal to the curve at this point and determine where it intersects the y-axis.
$\textbf{Step 1: Find the slope of the tangent line}$
The derivative of $y = \sqrt{x}$ is: $$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$$ At $x = 25$, the slope of the tangent is: $$\frac{dy}{dx} = \frac{1}{2\sqrt{25}} = \frac{1}{10}$$ Thus, the slope of the tangent line at $(25, 5)$ is $\frac{1}{10}$.
$\textbf{Step 2: Find the slope of the normal line}$
The slope of the normal line is the negative reciprocal of the slope of the tangent. So, the slope of the normal line is: $$m_{\text{normal}} = -10$$
$\textbf{Step 3: Equation of the normal line}$
The equation of the normal line can be written as: $$y - y_1 = m(x - x_1)$$ Substituting $m = -10$, $x_1 = 25$, and $y_1 = 5$: $$y - 5 = -10(x - 25)$$ Simplifying: $$y - 5 = -10x + 250$$ $$y = -10x + 255$$ \textbf{Step 4: Find the y-intercept} To find the point where the normal line intersects the y-axis, set $x = 0$: $$y = -10(0) + 255 = 255$$ Thus, the normal line intersects the y-axis at $(0, 255)$.

The correct option is (B) : $(0,\ 255)$