Question

whats is the equation of the tangent to the curve y - 5x*2 = 0 and parallel to line 5x + y + 1 = 0​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Equation of the curve is}$

$\mathsf{y=5\,x^2}$

$\underline{\textbf{To find:}}$

$\textsf{The equation of the tangent parallel to 5x+y+1=0}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,\;y=5x^2}$

$\textsf{Differentiate with respect to x}$

$\mathsf{\dfrac{dy}{dx}=10\,x}$

$\implies\textsf{Slope\;of\;tangent=10\,x}$

$\textsf{But the tangent is parallel to 5x+y+1=0}$

$\therefore\textsf{Their slopes are equal}$

$\implies\textsf{Slope of tangent=Slope of 5x+y+1=0}$

$\implies\mathsf{10x=-5}$

$\implies\mathsf{x=\dfrac{-5}{10}}$

$\implies\mathsf{x=\dfrac{-1}{2}}$

$\mathsf{when\;x=\dfrac{-1}{2},\;y=5\left(\dfrac{-1}{2}\right)^2}$

$\implies\mathsf{y=5\left(\dfrac{1}{4}\right)}$

$\implies\mathsf{y=\dfrac{5}{4}}$

$\implies\mathsf{Tangential\;point\;is\left(\dfrac{-1}{2},\dfrac{5}{4}\right)}$

$\textsf{The equation of the required tangent is}$

$\mathsf{y-y_1=m(x-x_1)}$

$\mathsf{y-\dfrac{5}{4}=-5\left(x+\dfrac{1}{2}\right)}$

$\mathsf{\dfrac{4y-5}{4}=-5\left(\dfrac{2x+1}{2}\right)}$

$\mathsf{\dfrac{4y-5}{2}=-5(2x+1)}$

$\mathsf{4y-5=-10(2x+1)}$

$\mathsf{4y-5=-20x-10}$

$\mathsf{20x+4y+5=0}$

$\implies\boxed{\mathsf{5x+y+\dfrac{5}{4}=0}}$