Question

Find the equation of
tangent and normal at point
(a, a) of the parabola
y (2a-x) = x²​

Answer 1

Answer:

Refer to the attachment :)

Answer 2

$\huge\boxed{\fcolorbox{red}{yellow}{your \: \: answer}}$

Given Curve is :-

$y {2}^{} = 4ax$

We need to find equation of target and normal at

We need to find equation of target & normal at

$(a {t}^{2} , 2at)$

We know that,

Slope of tangent is

$\frac{dy}{dx}$

${y}^{2} = 4ax$

Differentiating w.r.t.x

$\frac{d( {y}^{2}) }{dx} = \frac{d(4ax)}{dx}$

$\frac{d ({y}^{2}) }{dx} \times \frac{dy}{dy} = 4a \frac{d(x)}{dx}$

$= (answer)$

Hope it's help u.......

$<marquee>$

❣️Stay home and stay safe❣️

✍️-Answer by brainly queen✍️

♥️Be happy always♥️