
Equation of tangent to parabola
Equation of tangent to parabola $ y^2\,=\,4ax $ ant any parametric coordinate p(t)=$p(\,at^2,\,2at)$ where t is a parameter and $t\,\in R $ is given by $ yt\,=\,x\,+\,at^2 $ .
Hence 1/t is the slope of tangent at point P(t).
Let m=1/t Hence equation of tangent will be $\frac{y}{m}\,=\,x\,+\,\frac{a}{m^2} $
i.e tangent is y=mx + a/m where m is the slope of the tangent.
Application of tangent in slope form
Find the tangent to parabola $y^2\,=\,8x $ drawn from the point (-1,-1)
We know that equation of tangent with slope m to the parabola $ y^2\,=\,4ax $ is given by y=mx+a/m .
Here a=2 Hence y=mx+2/m is tangent with slope m to parabola $y^2\,=\,8x$ , And tangent passes through the point (-1,-1).
Hence we get -1=-m+2/m ; $m^2\,-m\,-2\,=0 $
Hence m = 2 , -1 are slope of required tangents.
Thuse Equation of tangents to the parabola $y^2\,=\,8x$ is y=2x+1 , and y=-x-2.
If equation of parabola is $ (y-k)^2\,=\,4a(x-h) $ then equation of tangent in slope for will be (y-k) = m(x-h) + a/m
Similarly if equation of parabola is $x^2=4ay $ then tangent in slope form is y=mx – $am^2$ and for parabola of form $ (x-h)^2\,=\,4a(y-k) $ tangent with slope m is $y-k=m(x-h)-am^2 $