Check the solution to "Line x+ky+k2=0 is tangent to curve y2=4x at..."

siliciooy0j

Answered question

2022-03-24

Line

x + k y + k^{2} = 0

is tangent to curve

y^{2} = 4 x

at point P. Find the coordinates of P in terms of k.

Jared Kemp

Beginner2022-03-25Added 14 answers

Step 1
We are told that

x = - k y - k^{2}

is tangential to

x = {\frac{14}{y}}^{2}

We have

\frac{d x}{d y} = \frac{24}{y}

\frac{2}{4} y = - k

Hence

y = - 2 k

and

x = \frac{1}{2} {(- 2 k)}^{2} = k^{2}

kattylouxlvc

Beginner2022-03-26Added 11 answers

Step 1
The line

x + k y + k^{2} = 0

is tangent to

y^{2} = 4 x

, which is a parabola .
Substituting

x = \frac{y^{2}}{4}

, in the equation of the given line, we get

y^{2} + 4 k y + k^{2} = 0

.
Now as the line is a tangent, there is only one point of intersection, hence this quadratic equation has equal roots, therefore the determinant

(b^{2} - 4 a c) = 0

.
Now by quadratic formula. Roots

= \frac{- b + \sqrt{D}}{2 a}

\frac{- b - \sqrt{D}}{2 a}

, but as

D = 0

, both roots are

\frac{- b}{2 a}

, which is equal to

\frac{- 4 k}{2} = - 2 k

. Therefore,

y = - 2 k

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