The line which is parallel to the x-axis and | vedclass

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(B) The equation of the curve is $y = \sqrt{x}$.On differentiating with respect to $x$,the slope of the tangent at any point $(x_1, y_1)$ is given by

Vedclass · Accepted Answer

$y = \frac{1}{2}$

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(B) The equation of the curve is $y = \sqrt{x}$.On differentiating with respect to $x$,the slope of the tangent at any point $(x_1, y_1)$ is given by $\frac{dy}{dx} = \frac{1}{2\sqrt{x_1}}$.Let the line parallel to the $x$-axis be $y = k$. The slope of this line is $m_2 = 0$.The angle between the curve and the line is $45^{\circ}$.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$	an 45^{\circ} = \left| \frac{\frac{1}{2\sqrt{x_1}} - 0}{1 + (\frac{1}{2\sqrt{x_1}})(0)} ight|$$1 = \frac{1}{2\sqrt{x_1}}$$2\sqrt{x_1} = 1 \Rightarrow \sqrt{x_1} = \frac{1}{2}$.Since $y_1 = \sqrt{x_1}$,we get $y_1 = \frac{1}{2}$.Since the line is $y = k$ and it passes through the point where $y = \frac{1}{2}$,the equation of the line is $y = \frac{1}{2}$.

The line which is parallel to the $x$-axis and crosses the curve $y = \sqrt{x}$ at an angle of $45^{\circ}$ is:

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