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How to find equation of tangent to parabola at parametric , slope form

Equation of tangent to parabola

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 $

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