Question

( x^{2}-y^{2}-8 x+2 y+11=0 )
(2,1) tangent
Slope of tongent at (2,1) is
[
begin{array}{l}
f^{prime}(2,1)=2 x+2=8
(x-4)^{2}-(y-1)^{2}=4
quad(y-1)^{2}=(x-4-2)(x-4+2)
f(x)=y=1 pm sqrt{(x-6)(x-2)}
f^{prime}(x)=pm frac{1(2 x-8)}{2 sqrt{(x-6)(x-2)}}
frac{y^{prime}(2)}{x-2}=frac{1}{0}
end{array}
]
( x=2 )

# 1. The equation of the tangent to the conic x2 - y2 – 8x + 2y + 11 = 0 at (2, 1), is (a) x + 2 = 0 (b) 2x + 1 = 0 (C) x – 2 = 0 (d) x + y + 1 = 0

Solution