Question

If the equation of the tangent line to the curve $y=\sqrt{5x-3}-2$ which is parallel to the line $4x-2y+3=0$ is $ax-40y-103=0$. Find $a$

Answer 1

$y=\sqrt{5x-3}-2$

$\frac{dy}{dx}=\frac{1}{2\sqrt{5x-3}}\times 5-0$

$m_1=\frac{dy}{dx}=\frac{5}{2\sqrt{5x-3}}$

$4x-2y+3=0$

$m_2=\frac{4}{2}=2$

$m_1=m_2$ lines are parallel

$\frac{5}{2\sqrt{5x-3}}=\frac{2}{1}$

$5=4\sqrt{5x-3}$

$25=16(5x-3)$

$25=80x-48$

$73=80x$

$x=\frac{73}{80}$

$y=\sqrt{\frac{5\times 73}{80}-3}-2$

$=\sqrt{\frac{25}{16}}-2$

$=-\frac{3}{4}$

Equation of the tangent pt. $(\frac{73}{80},\frac{-3}{4})$

slope = 2

$y-y_1=m(x-x_1)$

$y+\frac{3}{4}=2(x-\frac{73}{80})$

$40y+30=80x-73$

$80x-40y-103=0$