 Hello and how are you all today? The question says, find the equation of the tangent and normal to the parabola y square equal to 4ax at the point 80 square to 80. Here we are given the equation of the parabola as y square equal to 4ax on differentiating with respect to x we get 2y dy by dx is equal to 4a or we have dy by dx equal to 4a upon 2y we have dy by dx on simplifying it we have 2a by y right. Now this is the slope of the tangent points 80 square to 80 will be 2a upon y that is 280 on simplifying it we have the slope as 1 upon t right. We have the slope of the tangent as 1 upon t so slope of the normal which we know is minus 1 upon dy by dx will be minus t. Now we have the slope of the tangent as well as the normal so let us find the equation of slope equation of as well as equation of norm we have y minus y1 which is 280 equal to 1 upon t x minus x1 which implies dy minus 280 square equal to x minus 80 square which is further equal to dy is equal to x plus 80 square. Now here also we have y minus 280 equal to minus t x minus 80 square which implies y minus 280 equal to tx minus 80 cube which is equal to that is y is equal to here we had minus t sorry so it will be minus tx plus 280 plus 80 cube right. So the answer to the given question is equation of the tangent is ty equal to x plus 80 square and equation of the normal is y is equal to minus tx plus 280 plus 80 cube right. This completes our session hope you understood the concept well to take care of your calculations while calculating the slope of tangent as well as the moment and have a nice day.