 Hi, and welcome to our session. Let us discuss the following question. The question says proof that the product of the lens of the perpendicular stront from the points squared root of a squared minus b squared 0 and minus squared root of a squared minus b squared 0 to the line x by a cos theta plus y by b sin theta is equal to 1s b squared. Let's now begin with the solution. Let p1 be the length of the perpendicular from the point squared root of a squared minus b squared 0 to the line x by a cos theta plus y by b sin theta is equal to 1. We know that the perpendicular distance of the line a x plus b y plus c is equal to 0 from a point x and y1 is given by mod of a x1 plus b y1 plus c upon square root of a squared plus b squared a is equal to cos theta by a b is equal to sin theta by b and c is equal to minus 1. And point x1 y1 is square root of a squared minus b squared 0. So now, substitute these values in this formula. And this distance is equal to p1. So we have p1 is equal to mod of cos theta by a into square root of a squared minus b squared plus sin theta by b into 0 minus 1 upon square root of cos squared theta by a squared plus sin squared theta by b squared. This is equal to square root of a squared minus b squared by a into cos theta minus 1 upon square root of cos squared theta upon a squared plus sin squared theta upon b squared. Let's name this equation as equation number 1. Now, let p2 be the length of perpendicular point minus square root of a squared minus b squared 0 to the line a cos theta plus y by b sin theta is equal to 1. By using the formula for perpendicular distance, we find that p2 is equal to minus square root of a squared minus b squared upon a into cos theta plus 0 minus 1 upon square root of cos squared theta a squared plus sin squared theta upon b squared. Let's name this equation as equation number 2. We have to show that product of these two perpendicular is b squared, that is, p1 into p2 is equal to b squared. So let's now find p1 into p2. p1 is equal to square root of a squared minus b squared upon a into cos theta minus 1 upon square root of cos squared theta upon a squared plus sin squared theta upon b squared p2 is equal to minus square root of a squared minus b squared by a into cos theta minus 1 upon cos squared theta upon a squared plus sin squared theta upon b squared in square root. This is equal to minus square root of a squared minus b squared by a cos theta minus 1 into square root of a squared minus b squared upon a cos theta plus 1 upon cos squared theta upon a squared plus sin squared theta upon b squared. And this is equal to minus a squared minus b squared upon a squared into cos squared theta minus 1 upon b squared cos squared theta plus a squared sin squared theta upon a squared b squared. And this is equal to minus a squared cos squared theta minus b squared cos squared theta minus a squared upon a squared upon b squared cos squared theta plus a squared sin squared theta upon a squared b squared. This is equal to minus a squared cos squared theta plus b square cos square theta plus a square upon a square by b square cos square theta plus a square sine square theta by a square b square. By taking a square common from these two terms, we get a square into 1 minus cos square theta and this is equal to a square sine square theta. So now we have a square sine square theta plus b square cos square theta upon a square upon b square cos square theta plus a square sine square theta upon a square b square and this is equal to a square b square upon a square and this is equal to b square. Hence, we have proved that product of lengths of perpendicular stront from the point square root of a square minus b square 0 and minus square root of a square minus b square 0 to the line x by a cos theta plus y by b sine theta is equal to 1 s b square. This completes the creation. Bye and take care.