Solve the problem.
-Write an equation for the tangent line to the curve $y ^ { 2 } - x = 4$ at the point $( 1 , \sqrt { 5 } )$.
A) $x - 2 y + 9 = 0$
B) $x - 2 \sqrt { 5 } y + 1 = 0$
C) $x - 2 \sqrt { 5 } y + 9 = 0$
D) $x - 2 y + 1 = 0$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the curve $y^2 - x = 4$ at the point $(1, \sqrt{5})$, we first differentiate the given equation implicitly with respect to $x$:$$2y\frac{dy}{dx} - 1 = 0 \Rightarrow \frac{dy}{dx} = \frac{1}{2y}.$$Substituting $y = \sqrt{5}$ into the derivative gives the slope of the tangent line:$$\frac{dy}{dx} = \frac{1}{2\sqrt{5}}.$$The slope-intercept form of a line is $y = mx + b$, where $m$ is the slope. However, for the tangent line equation, we use the point-slope form, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point of tangency and $m$ is the slope. Substituting $m = \frac{1}{2\sqrt{5}}$, $x_1 = 1$, and $y_1 = \sqrt{5}$, we get:$$y - \sqrt{5} = \frac{1}{2\sqrt{5}}(x - 1).$$Multiplying both sides by $2\sqrt{5}$ to clear the fraction and rearrange to the general form $Ax + By + C = 0$, we get:$$x - 2\sqrt{5}y + 9 = 0.$$Therefore, the correct answer is C.

Solve the Problem y2−x=4y ^ { 2 } - x = 4y2−x=4

Solve the Problem $y ^ { 2 } - x = 4$