Question

Find the equation of normal to parabola x^2=4y at 6,9

Answer 1

Answer:

x + 3y = 33

Step-by-step explanation:

Given equation of the parabola is x² = 4y

=> Slope of the tangent at any point  is

2x = 4(dy/dx)

=> dy/dx = x/2

Hence Slope of tangent at P(6, 9) will be 6/2 = 3.

Hence Slope of Normal will be -1/3.

We now the slope of the normal and point (6, 9) at which normal is drawn

So, equation of the normal would be

y - 9/x - 6 = -1/3

=>3y - 27 = -x + 6

=> x + 3y = 33 is the required equation of the normal at (6, 9).

Hope, it helped !

Answer 2

Answer:

x+3y-33=0

Step-by-step explanation:

Concept:  The equation of any line perpendicular to ax+by+c=0 is of the form

bx-ay+k=0  

Equation of the given parabola is  x^2=4y  The equation of tangent at $(x_1,y_1)$ is of the form

$xx_{1}=4(\frac{y+y_1}{2}) \\xx_{1}=2(y+y_1) \\\\But\:(x_1,y_1)=(6,9)$ 6x=2(y+9) 6x=2y+18 6x-2y-18=0 3x-y-9=0  

The equation of normal is of the form -x-3y+k=0 It passes through (6,9) -6-3(9)+k=0 -6-27+k=0 k=33  The equation of normal is -x-3y+33=0 x+3y-33=0