Question

Find the equation of line normal to x^3+3y^2=3xy^2 at the point (1,1)

Answer 1

Answer:

Solution :

The given line is x+3y=4ory=−13x+43

∴ slope of the given line =−13

So, the slope of the required normal =−13...(i)

Let the point of contact be (x1,y1)

ow, 3x2−y2=8⇒6x−2y.dydx=0 [on differentiating w.r.t..x]

⇒dydx=3xy⇒(dydx)(x1,y1)=3x1y1

∴ slope of the normal =−1(dydx)(x1,y1)=−y13x1 ..(i)

Thus, from (i) and (ii) we get −y13x1=−13⇒y1=x1

Also, since (x1,y1) lies on the given curve, we have

3x21−y21=8or3x21−x21=8 [∵y1=x1]

or x21=4orx1=±2

∴y1=±2 [∵y1=x1]

Thus, the point of contact are (2, 2) and (−2,−2).

So, the equation of the required normal at (2,2) is y−2x−2=−13

i.e., x+3y−8=0

The equation of the required normal at (−2,−2) is y+2x+2=−13,

i.e., x+3y+8=0