 Hi and welcome to the session. My name is Priyanka and the question says, find the equations of the tangent and normal to the hyperbola x square upon a square minus y square upon b square is equal to 1 at the point x0, y0. So here we are given equation of the hyperbola as x square upon a square minus y square upon b square equal to 1. Now on differentiating the given equation with respect to x we get 2x upon a square minus 2y upon b square dy by dx equal to 0. That implies now dividing the whole equation by 2 we are left with x upon a square minus y upon b square dy by dx equal to 0. That further implies x upon a square is equal to y upon b square dy by dx. We have the value of dy by dx as b square x upon a square y. Now the value of this dy by dx at the given points that is given to us as x0, y0 will be equal to x0, b square upon y0, a square isn't it? Now since the value of dy by dx at x0, y0 is this. That means this is the slope of the tangent. Since slope of the tangent is this, so slope of the normal is reciprocal to 8 having a minus sign. This is the slope of the tangent and slope of the normal. Since we need to find the equation of the tangent and normal we need to have the slope for both tangent and normal. Now let's proceed by finding the equation of tangent as well as equation of normal. Now here we have y minus y1 that is y0 equal to slope that is x0, b square upon y0, a square x minus x0. Which implies y, y0, a square minus y0, square, a square equal to x, x0, b square minus x0, square, b square. This further gives us on dividing both the sides by x square, sorry, a square, b square we get y, y0, a square minus y0, square, a square divided by a square, b square. And on the other side also this is equal to y0 upon b square minus y square upon b square equal to x, x0 upon a square minus x square upon a square. But we know that x square upon a square minus y square upon b square is equal to 1 and at points x0, y0. So we have, so we have this equal to 1. So this means that x, x0 upon a square minus y, y0 upon b square will be equal to 1. Now this is the equation of the tangent. Now in the same manner we will be finding out equation of the normal. It will be y minus y0 equal to slope of the normal that is found out above as minus a square y0 upon b square x0, x minus x0. This implies b square x0, y minus b square x0, y0 equal to minus a square y0, x plus a square y0, x0. Now this gives us on taking b square x0 common we are left with y minus y0 and here on taking minus a square y0 we are left with x minus x0. Which can be now written as y minus y0 upon a square y0 plus x minus x0 upon b square x0 is equal to 0. This is equation of the normal. So let us write down both the answers we have for tangent. It is x, x0 upon a square minus y, y0 upon b square equal to 1. And equation of the normal is y minus y0 upon a square y0 plus x minus x0 upon b square x0 is equal to 0. So this completes the session. Hope you understood the whole process when you enjoyed it too. Have a nice day ahead.