 Hi and welcome to the session. Let us discuss the calling question. The question says find the point on the curve y squared equals to 4x which is nearest to the point o minus 8. We begin with the solution. Let the required point be p having coordinates x by y minus. Given curve as y squared equals to 4x, question this point lies on the curve y squared equals to 4x. So as the curve, sorry, as the point p having coordinates x1 by 1 lies on y squared equals to 4x. Therefore, y1 squared is equal to 4x1. Now, equation of tangent y squared equals to 4x at point x1 by 1 is given by minus y1 equals to 2 by y1 into x minus x1. Now, equation of line through x1 by 1 minus x1 equals to minus 8 minus y1 divided by 2 minus x1. Let's name this equation as 1 and this as 2. Now, from 1, we have y minus y1 divided by x minus x1 equals to 2 by y1. Substituting in place of y minus y1 divided by x minus x1, we get y1 equals to minus 8 minus y1 divided by 2 minus x1. This implies 4 minus 2x1 is equal to minus 8 y1 plus, sorry, minus y1 squared and squared plus 8 y1 plus 4 minus 2x1 is equal to 0. Let's name this equation as equation number 3. We know that y1 squared is equal to 4x1. y1 squared by 2 is equal to 2 formula. We find that y by 8 plus minus square root of 224 divided minus 2 into square root of 24x1. 2 into square root of 40. x1 is equal to minus 8 plus minus 2 into square root of 40 whole squared divided by minus 8 plus minus 2 into square root of 40 whole squared divided by 4 by coordinate is minus 8 plus minus into square root of 40. So, this is our required answer. So, this completes this session. I am taking care.