Question

Find the equation of tangent to the curve $y=\sqrt{5x-3}-2$, which is parallel to the line $4x-2y+3=0$?

• (A) $80x-40y+103=0$
• (B) $80x+40y+103=0$
• (C) $80x+40y-103=0$
• (D) $80x-40y-103=0$

Answer 1

The Correct Option is D

Explanation:
Given: Equation of curve is $y=\sqrt{5x-3}-2$ and the tangent to the curve $y=\sqrt{5x-3}-2$ is parallel to the line $4x-2y+3=0$The given line $4x-2y+3=0$ can be re-written as:$y=2x+\left(\frac{3}{2}\right)=0$Now by comparing the above equation of line with $y=mx+c$ we get,$m=2$ and $c=\frac{3}{2}$$\because$ The line $4x-2y+3=0$ is parallel to the tangent to the curve $y=\sqrt{5x-3}-2$As we know that if two lines are parallel then their slope is same.So, the slope of the tangent to the curve $y=\sqrt{5x-3}-2$ is $m=2$Let, the point of contact be $(x_1,y_1)$As we know that slope of the tangent at any point say $(x_1,y_1)$ to a curve is given by:$m={\left[\frac{dy}{dx}\right]}_{(x_1,y_1)}$$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\cdot \frac{1}{\sqrt{5x-3}}\cdot 5-0=\frac{5}{2\sqrt{5x-3}}$$\Rightarrow {\left[\frac{dy}{dx}\right]}_{(x_1,y_1)}=\frac{5}{2\sqrt{5x_1-3}}$$\because$ Slope of tangent to the curve $y=\sqrt{5x-3}-2$ is $m=2$$\Rightarrow 2=\frac{5}{2\sqrt{5x_1-3}}$By squaring both the sides of the above equation we get:$4=\frac{25}{4\cdot (5x_1-3)}$$\Rightarrow x_1=\frac{73}{80}$$\because (x_1,y_1)$ is point of conctact i.e., $(x_1,y_1)$ will satisfy the equation of curve:$y=\sqrt{5x-3}-2$$\Rightarrow y_1=\sqrt{5x_1-3}-2$By substituting $x_1=\frac{73}{80}$ in the above equation we get:$y_1=-\frac{3}{4}$So, the point of contact is: $(\frac{73}{80},-\frac{3}{4})$As we know that equation of tangent at any point say $(x_1,y_1)$ is given by:$y-y_1={\left[\frac{dy}{dx}\right]}_{(x_1,y_1)}\cdot (x-x_1)$$\Rightarrow y+\frac{3}{4}=2\cdot (x-\frac{73}{80})$$\Rightarrow 80x-40y-103=0$So, the equation of tangent to the given curve at the point $(\frac{73}{80},-\frac{3}{4})$ is $80x-40y-103=0$Hence, the correct option is (D).