Question

Let a be the point where the curve 5a^2x^3+10ax^2+x+2y-4 meeys y-axis. then the equation of tangent to the curve at the point a meets the curve again is?

Answer 1

$\textbf{Given:}$

$\text{Equation of the curve is 5a^2\,x^3+10a\,x^2+x+2y-4=0 }$

$\textbf{To find:}$

$\text{Equation of the tangent at the point where the curve meets y axis}$

$\textbf{Solution:}$

$5a^2\,x^3+10a\,x^2+x+2y-4=0$

$\text{To find the points where the curve meets y axis, put x=0}$

$\implies\,2y-4=0$

$\implies\,y=2$

$\therefore\textbf{(0,2) is the required point at which the curve meets y axis}$

$5a^2\,x^3+10a\,x^2+x+2y-4=0$

$\text{Differentiate with respect to x}$

$5a^2(3x^2)+10a(2x)+1+2\,\dfrac{dy}{dx}=0$

$15a^2\,x^2+20a\,x+1+2\,\dfrac{dy}{dx}=0$

$2\,\dfrac{dy}{dx}=-(15a^2\,x^2+20a\,x+1)$

$\dfrac{dy}{dx}=\dfrac{-(15a^2\,x^2+20a\,x+1)}{2}$

$\textbf{Slope of tangent}$

$m=(\dfrac{dy}{dx})_{(0,2)}$

$m=\dfrac{-(15a^2\,(0)^2+20a\,(0)+1)}{2}$

$m=\dfrac{-1}{2}$

$\textbf{The equation of tangent is}$

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

$y-2=\frac{-1}{2}(x-0)$

$y-2=\frac{-x}{2}$

$2y-4=-x$

$\implies\bf\,x+2y-4=0$

$\therefore\textbf{The required tangent is x+2y-4=0}$

Find more:

Equation of the tangent to the curve y = 1-e^-x/2 at the point where the curve cuts y axis is

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