 Hi, and how are you all today? My name is Priyanka and the question says find the points on the curve y is equal to x q at which the slope of the tangent is equal to the y coordinate of the points. Here we are given the equation the curve as y is equal to x q on differentiating y with respect to x we get dy by dx equal to 3x square, right? Now let the point on the curve be x1 y1 right, so at this point dy by dx at x1 y1 is equal to 3x1 square which is also the slope of the tangent. So at x which is x1 y is equal to x1 q, right? Now we are given question that of tangents are equal to the y coordinate that is 3x1 square is equal to x1 cube, right? That implies 3x1 square minus x1 cube is equal to 0 taking x1 square common. We are left with 3x1 minus sorry, taking x1 square common. We are left with 3 minus x1 equal to 0. So that means either x1 square is equal to 0 or 3 minus x1 is equal to 0. That implies that x1 is equal to 0 or x1 is equal to 3. Now when x1 is equal to 0 the value of y1 is equal to 0 and when x1 is equal to 3 value of y1 is equal to 3 cube that is equal to 27, hence the points are 0 comma 0 and 3 comma 27. Wait, this is the answer to this part. Hope you understood it. Sorry, this is the answer to this question. Hope you understood it well and enjoyed it too. Have a nice day.