Greetings Aspirant,

The Answer to your question is as follow:

Let the equation of the tangent be

y=mx+c

Since it passes through (2,8) hence

8=2m+c

Or

c=8−2m

Hence the equation of the tangent becomes

y=mx−2m+8

Substituting in the equation of the hyperbola

5x(Square) −y(Square) =5

5x(Square) − (mx−2m+8)(Square) = 5

5x(Square)−5−(m(Square)x(Square)+4m(Square)+64−4m(Square)x−32m+16mx)=0

x(Square)(5−m(Square) )+x(4m(Square) −16m)−4m(Square) +32m−69=0

Since it is a tangent hence D is 0.

Or

B(Square) −4AC=0

(4m(Square) −16m)(Square) −4(5−m(Square) ) (−4m(Square) +32m−69)=0

m=3 and m=23/3

Hence

y=3x+2 and 3y=23x−22

For Your Better Understanding i have also uploaded an image of the same answer to your question. You may go through it.

Answer Image (https://drive.google.com/drive/folders/17Q4BTcoBKGmANQIhhdKF_5ZZkJcqGWAP?usp=sharing)

Hope My Answer Helped You.

Best of Luck For Future.