 Hello and welcome to the session. In this session, we will discuss tangents and normals. We will see how we use differentiation to find the equation of the tangent line and the normal line to a curve at a given point. Equation of the tangent at the point x naught, y naught to the curve y equal to fx is given by y minus y naught equal to dy by dx at the point x naught y naught into x minus x naught. Now, dy by dx at the point x naught y naught is the slope of the tangent to the curve y equal to fx. Let this be denoted by f dash x naught. So we have equation of the tangent is given by y minus y naught equal to f dash x naught into x minus x naught. Then again, if a tangent line to the curve y equal to fx makes an angle theta with x axis in the positive direction, then dy upon dx that is the slope of the tangent is equal to tan theta. If the slope of the tangent line is 0 that is we have tan theta is equal to 0. This implies theta is equal to 0. So this means that the tangent line is parallel to x axis and in this case the equation of the tangent at the point x naught y naught is given by y equal to y naught. If we have theta tends to pi by 2 then we have tan theta tends to infinity which means that the tangent line is perpendicular to the x axis that is it is parallel to y axis and in this case the equation of the tangent at the point x naught y naught is given by x equal to x naught. Let's try and find out the equation of the tangent to the curve given by y equal to x to the power 4 minus 6x cube plus 13x square minus 10x plus 5 at the point 1,3. From the point 1,3 we have x naught is 1 and y naught is 3. Now we find out dy by dx from this equation of the curve. We get dy by dx is equal to 4x cube minus 18x square plus 26x minus 10. Then dy by dx that is the slope of the tangent at the point 1,3 is given by 4 into 1 cube minus 18 into 1 square plus 26 into 1 minus 10 and that comes out to be equal to 2. Then the required equation of the tangent at the point x naught y naught that is 1,3 is given by y minus y naught equal to dy by dx at x naught y naught that is equal to 2 into x minus x naught. This comes out to be equal to 2x minus y plus 1 equal to 0. This is the equation of the tangent. Next we have equation of the normal to the curve y equal to fx at a point x naught y naught is given by y minus y naught equal to minus 1 upon dy by dx at the point x naught y naught into x minus x naught. This is the slope of the normal that is minus 1 upon dy by dx at the point x naught y naught is the slope of the normal. Since the normal is perpendicular to the tangent and slope of the tangent is dy upon dx at x naught y naught so this becomes the slope of the normal. Let this be equal to minus 1 upon f dash x naught if f dash x naught is not equal to 0. So we have equation of the normal is y minus y naught into f dash x naught plus x minus x naught equal to 0. If we have dy by dx at the point x naught y naught that is f dash x naught is 0 then the equation of the normal is given by x equal to x naught. And if dy by dx at the point x naught y naught that is f dash x naught does not exist then the normal is parallel to x axis. Then the equation of the normal is given by y equal to y naught. Let's consider the same example that we have considered above in which the curve is given by y equal to x to the power 4 minus 6x cube plus 13x square minus 10x plus 5 and the point given is 1 comma 3. Now from here we have dy by dx at the point 1 comma 3 or you can say f dash x naught is equal to 2 so equation of the normal is given by y minus y naught into f dash x naught plus x minus x naught equal to 0. That is we have x plus 2y minus 7 equal to 0 is the required equation of the normal. This completes the session. Hope you have understood tangents and normals and how we find the equation of the tangents and equation of the normals at given point.