Question

Find k if y + kx = 5 is a tangent to the curve y2 = 4x

Answer 1

Answer:

Given

y = 5 - kx

Substitute the value for intersection points

$(5 - kx) {}^{2} = 4x \\ {k}^{2} {x}^{2} - (10k + 4)x + 25 = 0$

now if it touches then x is unique . In other words quadratic have equal roots which is possible if

D = 0

$(10k + 4) {}^{2} - 4(25) {k}^{2} = 0 \\ (10k + 4) {}^{2} - {(10k)}^{2} = 0 \\ (10k + 4 + 10k)(10k + 4 - 10k) = 0 \\ 20k =- 4 \\ k = \frac{-1}{5}$

Answer 2

Answer:

-1/5

Step-by-step explanation:

line : y = mx + c

curve : y2 = 4ax

condition for tangent : c = a/m

now....

line : y + kx = 5 i.e..y = -kx + 5

curve : y2 = 4x i,e..y2 = 4×1×x

hence....

5 = 1/-k

k = -1/5