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Consider hyperbola xy = 16 to find the following: (i) Coordinates of vertices (ii) Length of transverse axis (iii) Coordinates of foci (iv) Length of latus rectum (v) Equations of two directrices (vi) Equation of tangent at point (2, 8) (vii) Equation of normal at point (2, 8) (viii) Equation of chord of contact w.r.t. point (2, 3) (ix) Equation of chord which gets bisected at point (5, 6) (x) Equation of tangent having slope - 2 (xi) Equation of noraml having slope 2 |
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Answer» Solution :We havehyperbola 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` |
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