If the normals to the ellipse x2a2+y2b2=1 at the points x1, y1, x2, y2 and x3, y3 are concurrent, prove - Maths - Conic Sections

Question

If the normals to the ellipse x 2 a 2 + y 2 b 2 = 1   a t
  t h e   p o i n t s   x 1 ,   y 1 ,   x
2 ,   y 2   a n d   x 3 ,   y 3 are concurrent,
prove that
x 1 y 1 x 1 y 1 x 2 y 2 x 2 y 2 x 3 y 3 x 3 y 3 = 0

Rahul Raj · Accepted Answer

dear student

the equation of normal to the ellipsex2a2+y2b2=1 at the point (x1,y1) is(b2x1)y-(a2y1)x+(a2-b2)x1y1=0similarly other two normals at the points (x2,y2) and  (x3,y3)  will be(b2x2)y-(a2y2)x+(a2-b2)x2y2=0(b2x3)y-(a2y3)x+(a2-b2)x3y3=0as the three lines are collinear(b2x1)y-(a2y1)x+(a2-b2)x1y1=0(b2x2)y-(a2y2)x+(a2-b2)x2y2=0(b2x3)y-(a2y3)x+(a2-b2)x3y3=0we must have determinant of coefficients equal to zerob2x1a2y1(a2-b2)x1y1b2x2a2y2(a2-b2)x2y2b2x3a2y3(a2-b2)x3y3=0taking out common terms from each columnx1y1x1y1x2y2x2y2x3y3x3y3=0

regards

If the normals to the ellipse x 2 a 2 + y 2 b 2 = 1 a t t h e p o i n t s x 1 , y 1 , x 2 , y 2 a n d x 3 , y 3 are concurrent, prove that x 1 y 1 x 1 y 1 x 2 y 2 x 2 y 2 x 3 y 3 x 3 y 3 = 0