 Hi and welcome to the session my name is Priyanka and the question says find the equation of the tangent and normal to the given curves at the indicated points. Here we have y is equal to x raised to the power 4 minus 6x cube plus 13x square minus 10x plus 5 at 0 comma 5. Let us proceed with the solution. Here first of all we will rewrite the given equation of the curve and we will differentiate y with respect to x and I am doing so we get dy by dx is equal to 4x cube minus 18x square plus 26x minus 10. Right now we will find out the value of dy by dx at 0 comma 5 and it is minus 10 as the value of x in all the cases will be 0 so we are left with minus 10. So we have slope of tangent is dy by dx that is minus 10 and slope of normal would be minus 1 upon dy by dx that is minus 10 that will give us the value as 1 upon 10. Now let us find out equation of the tangent in one side equation of normal in the other side. We know that equation of tangent is y minus y1 equal to m that is slope bracket x minus x1 right and equation of normal is y minus y1 equal to 1 upon m sorry it is minus 1 upon m x minus x1. Now what we need to do is we will be substituting the values the points which are given to us is 0 comma 5. So we have y minus 5 equal to m that is minus 10 slope of the tangent x minus x1 that is 0. Here we have y minus 5 equal to minus 1 upon m is the slope of the normal that is 1 upon 10 x minus x1. Now let us simplify it. We have y minus 5 plus 10 x equal to 0 here and here we have 10 y minus 50 equal to which can be written as 10 x plus y is equal to 5 and here we can write it as x minus 10 y plus 50 equal to 0. So this is the equation for the tangent and this is equation for the normal right. So these two are the answers for the first part of the question. Hope you understood it well and enjoyed it too. Have a nice day ahead.