AM-1319

Line meeting a curve

A Math · Secondary 4 · Hard · Line meeting a curve

The equation of a curve is

y = kx^2 + 6x + 2k - 3

, where

k

is a constant. The line

y = 4x + 1

is tangent to the curve. Find the value of

k

.

$1 + \dfrac{\sqrt{6}}{2}$ or $1 - \dfrac{\sqrt{6}}{2}$
$1 + \dfrac{\sqrt{3}}{2}$ or $1 - \dfrac{\sqrt{3}}{2}$
$2 + \dfrac{\sqrt{6}}{2}$ or $2 - \dfrac{\sqrt{6}}{2}$
$\dfrac{4 + \sqrt{6}}{2}$ only

Sign in to check your answer and see the hint and the worked explanation. Free on PrepLab High.

Loading problem…