Question

Find the equation of the normal at the point ( am 2 , am 3 ) for the curve ay 2 = x 3 .

Answer 1

Equation of the curve is given as,

a y 2 = x 3

The slope of the tangent to the curve is given as,

slope =( dy dx )

Hence, the slope of the tangent of the given curve is given by differentiating with respect to x.

2ay( dy dx )=3 x 2 dy dx = 3 x 2 2ay

Slope of the tangent to the curve at ( a m 2 ,a m 3 ) is given by,

( dy dx ) ( a m 2 ,a m 3 ) = 3 ( a m 2 ) 2 2a( a m 3 ) = 3m 2

Slope of the normal is given as,

slope of the normal= −1 slope of the tangent

Hence, slope of the normal of the given curve is given by,

slope of the normal= −1 3 x 2 +2

Equation of the given line is,

x+14y+4=0 y= −1 14 x− 4 14

The above equation is written in the form y=mx+c, where m is the slope of the line.

Therefore, slope of the given line is −1 14 .

Normal is parallel to the given line; hence their slopes must also be same, therefore,

−1 3 x 2 +2 = −1 14 3 x 2 +2=14 3 x 2 =12 x 2 =4 x=±2

Coordinate of y when x=2 is,

y=8+4+6 =18

Coordinate of y when x=−2 is,

y=−8−4+6 =−6

The normal passes through points ( 2,18 ) and ( −2,−6 ).

Hence, equation of the normal through point ( 2,18 ) is,

y−18= −1 14 ( x−2 ) 14y−252=−x+2 x+14y−254=0

Equation of the normal through point ( −2,−6 ) is,

y−( −6 )= −1 14 ( x−( −2 ) ) y+6= −1 14 ( x+2 ) x+14y+86=0

Hence, the normal parallel to the given line is x+14y+86=0 and x+14y−254=0.