If the normal at four points `P_(i)(x_(i), (y_(i)) l, I = 1, 2, 3, 4` on the rectangular hyperbola `xy = c^(2)` meet at the point `Q(h, k),` prove that
`x_(1) + x_(2) + x_(3) + x_(4) = h, y_(1) + y_(2) + y_(3) + y_(4) = k`
`x_(1)x_(2)x_(3)x_(4) =y_(1)y_(2)y_(3)y_(4) =-c^(4)`

If the normals at four points P (x_i y_i), i = 1, 2, 3, 4 on the rectangular hyperbola xy = c^2 , meet at the point Q(h, k), then prove that x_1 + x_2 + x_3 + x_4 =h

If the normals at (x_(i),y_(i)) i=1,2,3,4 to the rectangular hyperbola xy=2 meet at the point (3,4) then

A circle cuts the rectangular hyperbola xy=1 in the points (x_(r),y_(r)), r=1,2,3,4 . Prove that x_(1)x_(2)x_(3)x_(4)=y_(1)y_(2)y_(3)y_(4)=1

If the normal at the point P(x_1y_1),i=1.2,3,4 on the hyperbola xy=c^2 are concurrent at the point Q(h, k), then ((x_1+x_2+x_3+x_4)(y_1+y_2+y_3+y_4))/(x_1x_2x_3x_4) is:

Tangents are drawn from the points (x_(1), y_(1))" and " (x_(2), y_(2)) to the rectanguler hyperbola xy = c^(2) . The normals at the points of contact meet at the point (h, k) . Prove that h [1/x_(1) + 1/x_(2)] = k [1/y_(1)+ 1/y_(2)] .

If the hyperbola xy=c^(2) intersects the circle x^(2)+y^(2)=a^(2)" is four points "P(x_(1),y_(1)), Q(x_(2),y_(2)), R(x_(3),y_(3)) and S(x_(4),y_(4)) then show that (i) x_(1)+x_(2)+x_(3)+x_(4)=0 (ii) y_(1)+y_(2)+y_(3)+y_(4)=0 (iii) x_(1)x_(2)x_(3)x_(4)=c^(4) (iv) y_(1)y_(2)y_(3)y_(4)=c^(4)

If the normals at two points (x_(1),y_(1)) and (x_(2),y_(2)) of the parabola y^(2)=4x meets again on the parabola, where x_(1)+x_(2)=8 then |y_(1)-y_(2)| is equal to

A triangle has vertices A_(i) (x_(i),y_(i)) for i= 1,2,3,. If the orthocenter of triangle is (0,0) then prove that |{:(x_(2)-x_(3),,y_(2)-y_(3),,y_(1)(y_(2)-y_(3))+x_(1)(x_(2)-x_(3))),(x_(3)-x_(1) ,,y_(3)-y_(1),,y_(2)(y_(3)-y_(1))+x_(2)(x_(3)-x_(1))),( x_(1)-x_(2),,y_(1)-y_(2),,y_(3)(y_(1)-y_(2))+x_(3)(x_(1)-x_(2))):}|=0

Evaluate : (i) (2x + 3y) ( 3x +4y)

If the circle x^2+y^2=a^2 intersects the hyperbola x y=c^2 at four points P(x_1, y_1),Q(x_2, y_2),R(x_3, y_3), and S(x_4, y_4), then x_1+x_2+x_3+x_4=0 y_1+y_2+y_3+y_4=0 x_1x_2x_3x_4=C^4 y_1y_2y_3y_4=C^4