Question

The equation of normal to x²/³+y²/³=a²/³ at(a/2√2,a/2√2) is.......,Select correct option from the given options.
(a) 2x+y=0
(b) y=1
(c) x=0
(d) x=y

Answer 1

The equation of the normal line is: y=−13x+73

Explanation:

Compute y'(x):

y'(x)=3x2

The slope, m, of the tangent line is at the point (1, 2) is y'(1):

m=y'(1)=3(1)2=3

The slope, n, of the normal to the curve is:

n=−1m=−13

Start with the slope-intercept form of the equation of a line:

y=mx+b

Substitute, 2 for y, −13 for m, 1 for x, and the solve for b:

2=−13(1)+b

b=73

The equation of the normal line is: y=−13x+73

Answer 2

Answer:

x=y

Step-by-step explanation:

The equation of the curve is $x^{\frac{2}{3} }+ y^{\frac{2}{3} }=a^{\frac{2}{3} }$ ....... (1)

Now, we have to calculate the equation of normal to the curve (1) at point ($\frac{a}{2\sqrt{2} }, \frac{a}{2\sqrt{2} }$).

First of all, we have to calculate the slope of the normal at that point.

Differentiating equation (1) with respect to x we get,

$\frac{2}{3} (\frac{1}{x^{\frac{1}{3} } } )+\frac{2}{3} (\frac{1}{y^{\frac{1}{3} } })\frac{dy}{dx} =0$

⇒ $\frac{dy}{dx} =-(\frac{y}{x} )^{\frac{1}{3} }$

⇒ -$\frac{dx}{dy} = (\frac{x}{y} )^{\frac{1}{3} }$  

⇒$[-\frac{dx}{dy} ]_{\frac{a}{2\sqrt{2} }, \frac{a}{2\sqrt{2} }} =1$.

Hence, the equation of the normal is

(y-a/2√2)=1(x-a/2√2)

⇒ x=y. (Answer)