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) : \((0,\ 255)\)

Answer 2

We are given the curve \(y = \sqrt{x}\) and the point (25, 5) on the curve. Our goal is to find the y-intercept of the normal line to the curve at this point.

First, we find the derivative of \(y\) with respect to \(x\): \[\frac{dy}{dx} = \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{\frac{1}{2}}) = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}\]

Next, we evaluate the derivative at the point (25, 5) to find the slope of the tangent line at that point: \[m_t = \frac{dy}{dx}\Big|_{x=25} = \frac{1}{2\sqrt{25}} = \frac{1}{2(5)} = \frac{1}{10}\]

The normal line is perpendicular to the tangent line at the point (25, 5). Therefore, the slope of the normal line, \(m_n\), is the negative reciprocal of the slope of the tangent line: \[m_n = -\frac{1}{m_t} = -\frac{1}{\frac{1}{10}} = -10\]

Now we have the slope of the normal line, \(m_n = -10\), and a point on the normal line, (25, 5). We can use the point-slope form of a line to find the equation of the normal line: \[y - y_1 = m_n(x - x_1)\] \[y - 5 = -10(x - 25)\]

Simplify the equation: \[y - 5 = -10x + 250\] \[y = -10x + 255\]

To find the y-intercept, we set \(x = 0\): \[y = -10(0) + 255 = 255\]

Therefore, the normal line intersects the y-axis at the point (0, 255).