 Hello and how are you all today? The question says find the equation of the tangent and normal to the given curves at the indicated point. Here we are given the equation of the curve as y is equal to x square at the indicated points are 0, 0. We are given y is equal to x square. On differentiating y with respect to x we get dy by dx is equal to 2x. Now the indicated points sorry the indicated point which was given to us is 0, 0. So the value of dy by dx at this point will be 2 into 0 which is equal to 0. That means slow of tangent which we take as m is equal to 0 and slow of normal which is minus 1 upon m will be equal to minus 1 upon 0. Now further equation of tangent is y minus y1 equal to slope x minus x1. So we have y equal to 0 as the equation of the tangent. Now equation of normal will be y minus y1 equal to slope of the normal x minus x1 which gives us 0 equal to minus x that further gives us x equal to 0. So the answer to this part is equation of the tangent as y is equal to 0 and to that of normal is x is equal to 0. This completes the session. Hope you find it interesting. Have a nice day.