00:02
Given that the point 3.
00:08
Minus 3, where 3 is the x coordinate and minus 3 is the y coordinate, the equation of curve is given to be y equal to x square plus 3.
00:23
First, we have to take the derivative of a given differential equation.
00:31
D differentiate with respect to x.
00:34
We get d .y by d x equal to d.
00:38
D .s.
00:39
X squared plus 3 simplify it d by d x x square plus d by d x 3 the derivative of constant term is 0 and the derivative of x square we get d y by d x equal to 2x it is basically represent the therefore it represents a slope of tangent at any given x coordinate at the point of tangency x1 is m1 equal to 2x the slope of the normal line is flat it be m2 m2 equal to minus 1 divided by m1 substitute the value of m1 we get minus 1 divided by 2 x the y coordinate y 1 at point of normalcy is y1 equal to x x2 plus 3 using slope point form of equation line y equal to minus 1 divided by 2 x1 x minus x1 plus x1 plus 3 now we have the normal line contain point 3 comma minus 3.
03:30
After substituting this value in above equation we get minus 3 equal to minus 1 divided by 2 x1 3 minus x1 plus x1 plus x1 plus 3 the value of x1 is nearly equal to 0 2 2 2 3 1 or x1 is nearly equal to minus 6 .7 to 3 .1.
04:07
We could take positive value.
04:22
The corresponding y -coordinate is y1 equal to x1 square plus 3.
04:40
Substitute the value of x1 which is 0 .2 whole square plus 3.
04:48
Simplify it we get 0 .04 .84 plus 3...