Question

Find the tangent at (_a,0) ;ay^2=x(x+a^2)

Answer 1

Answer:

sorry I don't know the answer

Answer 2

Answer:

Step-by-step explanation:

$\tt{Given\,\,curve,\,\,\purple{ay^2=x(x+a^2)}}$

To get the slope of the tangent, differentiating both sides w.r.t., x

$\sf{2ay\cdot\dfrac{dy}{dx}=(x+a^2)+x}$

$\sf{\implies\dfrac{dy}{dx}=\dfrac{2x+a^2}{2ay}}$

Slope of the tangent at $(-a,0)$ is,

$\sf{\implies\dfrac{dy}{dx}\bigg|_{(-a,0)}=\dfrac{2(-a)+a^2}{2a(0)}=\infty}$\

Equation of tangent:

$\rm{(y-0)=\dfrac{dy}{dx}\bigg|_{(-a,0)}\,(x+a)}$

$\rm{\implies(y-0)=(\infty)(x+a)}$

$\rm{\implies(y-0)=\dfrac{1}{0}\cdot(x+a)}$

$\rm{\implies0=(x+a)}$

$\rm{\implies\,x=-a}$

$\large{\sf{\blue{Hence,\,equation \,\,of\,\,the\,\,tangent\,\,at\,\,(-a,0)\,\,is\,\,x=-a}}}$