# circle and tangents , find length of intercept , tangents are drawn from external point

The tangents are drawn from the external point $(1,1)$ to the curve $$x^2-3y^2=1$$then find the length of intercept made by these tangents on the line $$x+y=1$$

• Welcome to MSE. We would like to see your work before answering questions. This is not a place for you to dump your homework questions – user507623 Dec 27 '17 at 17:45
• @psl people here can easily solve this question in less than a minute, but that's not the aim of MSE. Like what pippo said, we are not here to do homework for you. And that's exactly why Dr. Graubner is not giving you full solution. We want you to learn. – Karn Watcharasupat Dec 27 '17 at 18:05

set for the Tangent line $$y=mx+n$$ this line containes the Point $P(1;1)$ then our equation has the form $$y=mx+1-m$$ plug this equation in $$x^2-3y^2=1$$ and solve it for $x$ and compute $m$ Setting the discriminante equal to Zero doing this we obtain $$x^2(1-3m^2)+x(6m^2-6m)-3m^2+6m-4=0$$ and the discriminante is given by $$4-6m$$ can you finish now?
Here is a method that seems more intuitive to me. First write the gradient for the hyperbola. It represents the direction vector that is orthogonal to every point on the curve. The hyperbola is $f: x^2-3y^2=1$, so the gradient is $$\nabla f=\binom{2x}{-6y}$$ We know that the slope of any tangent lines will be perpendicular to the normals (gradient), so we can get the tangent slope by swapping the gradient terms and making one of them negative. There is a more formal way to explain this - maybe later. Anyway, the tangent line slopes will be $$slope=\binom{6y}{2x}$$ Now we can write a tangent line equation in parametric notation. $$\binom{x}{y}=\binom{1}{1}+t\binom{6y}{2x}$$ Of course, that represents two equations $$x=1+6ty\qquad\qquad y=1+2tx$$ Include the origianal hyperbola, $x^2-3y^2=1$ and we have 3 equation in 3 unknowns, x, y, and t. These solve quite easily to give two solutions, $$Point1=(-2,-1)\text{ and }Point2=(1,0)$$ With these tangent points, you can finish the solution for the line intersections.