Find the equation of tangent to the conic `x^2-y^2-8x+2y+11=0` at `(2,

1.	Find the equation of tangent to the conic `x^2-y^2-8x+2y+11=0` at `(2,1)`
Answer» Equation of the conic is , `x^2-y^2-8x+2y+11=0` Differentiating it with respect to `x`, `2x-2ydy/dx-8+2dy/dx = 0` `=>dy/dx(2-2y) = 8-2x` `=>dy/dx = (4-x)/(1-y)` At point `(2,1)`, `dy/dx = oo` `:.` slope is `oo`.Now, equation of line,`y = mx+c` At point `(2,1)`, `1 = 2m+c => c = 1-2m` `:.` Equation of tangent, `y = m(x)+(1-2m)` `=>y = m(x-2)+1` `=>(y-1)/m = (x-2)` As `m = oo`, `:. x-2 = 0` `=>x=2`, which is the required equation.

Discussion

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