We have hyperbola xy = 16.
(i) Clearly, transverse axis is `x-y=0` and conjugate axis is `x+y=0.`
Vertices are points of intersection of transverse axis `x-y=0` and hyperbola.
So, vertices are `A (4,4) and A'(-4,-4).`
(ii) Length of transverse axis is `"AA'"=8sqrt2=(=2a).`
(iii) Foci lie at distance 'ae' from centre on line `x-y=0.`
Now, `ae=4sqrt2sqrt2=8`
Therefore, foci are `F_(1)(4sqrt2, 4sqrt2) and F_(2)(-4sqrt2, -4sqrt2)`.
(iv) Length of latus rectum `=(2b^(2))/(a)=2a=8sqrt2`
(as a = b for rectangular hyperbola)
(v) Each of the directrices lies at distance `(a)/(e)(=4)` from centre and is parallel to the conjugate axis `x+y=0`.
So, equation of two directrices are `x+y=pm4sqrt2`.
(vi) Equation of tangent at point (2, 8) is
T = 0
`rArr" "(2y+8x)/(2)-16=0`
`rArr" "4x+y-16=0`
Alternatively, we can differentiate the equation of curve to get the slope of tangent.
Then, we get
`y+x(dy)/(dx)=0`
`therefore" "(dy)/(dx)=-(y)/(x)`
`therefore" "((dy)/(dx))_(("2,8"))=-(8)/(2)=-4`
So, equation of tangent at (2, 8) is
`y-8=-4(x-2)`
`rArr" "4x+y-16=0`
(vii) Equation of normal at (2,8) is `x-4y+30=0`.
(viii) Equation of chord of contact w.r.t. point (2, 3) is
T = 0
`rArr" "(2y+3x)/(2)-16=0`
`rArr" "3x+2y-32=0`
Equation of chord which gets bisected at point (5,6) is
T = `S_(1)`
`rArr" "(5y+6x)/(2)-16=(5)(6)-16`
`rArr" "6x+5y=60`
(x) To find the equation of tangent having slope `-2`, we differetiate the curve and compare derivative to `-2`.
`therefore" "(dy)/(dx)=-(y)/(x)=-2`
`therefore" "y=2x`
Solving this with hyperbola, we get points `P(sqrt8, 2sqrt8)`and `Q(-sqrt8,-2sqrt8)` on the hyperbola where slope of tangent is `-2.`
Equation of tangent at point P is
`y-2sqrt8=-2(x-sqrt8)`
`rArr" "2x+y=4sqrt8`
Equation of tangent at point Q is
`y+2sqrt8=-2(x+sqrt8)`
`rArr" "2x+y=-4sqrt8.`
(xi) We have `(dy)/(dx)=-(y)/(x)`
Thus, slope of normal at any point on the curve is `-(dx)/(dy)=(x)/(y)=2` (given)
`therefore" "x=2y`
Solving this with hyperbola, we get points `A(2sqrt8, sqrt8)` and `B(-2sqrt8-sqrt8)` on the hyperbola where slope of normal is 2.
Equation of normal at point A is
`y-sqrt8=2(x-2sqrt8)`
`rArr" "2x-y=3sqrt8`
Equation of normal at point B is
`y+sqrt8=2(x+2sqrt8)`
`rArr" "2x-y=-3sqrt8`
