We have the parametric equation #{(x=t^2+t-1),(y=2t^2-t+2):}#.
To show that #(-1,5)# lies on the curve defined above, we must show that there is a certain #t_A# such that at #t=t_A#, #x=-1,y=5#.
Thus, #{(-1=t_A^2+t_A-1),(5=2t_A^2-t_A+2):}#. Solving the top equation reveals that #t_A=0\ "or"\ -1#. Solving the bottom reveals that #t_A=3/2\ "or"\ -1#.
Then, at #t=-1#, #x=-1,y=5#; and therefore #(-1,5)# lies on the curve.
To find the slope at #A=(-1,5)#, we first find #("d"y)/("d"x)#. By the chain rule #("d"y)/("d"x)=("d"y)/("d"t)*("d"t)/("d"x)=("d"y)/("d"t)-:("d"x)/("d"t)#.
We can easily solve #("d"y)/("d"t)=4t-1# and #("d"x)/("d"t)=2t+1#. Thus, #("d"y)/("d"x)=(4t-1)/(2t+1)#.
At point #A=(-1,5)#, the corresponding #t# value is #t_A=-1#. Therefore, #[("d"y)/("d"x)]_(t=-1)=((4*-1)-1)/((2*-1)+1)=5#.
To find the line tangent to #A=(-1,5)#, recall the point-slope form of the line #y-y_0=m(x-x_0)#. We know that #y_0=5,x_0=-1,m=5#.
Substituting these values in shows that #y-5=5(x+1)#, or simply #y=5x+10#.
