# Thread: values-find word problem

1. ## values-find word problem

1. find the value of k for which y= kx+1 is a tangent to the curve y^2 = 8x

2. the hypotenuse of a triangle measures 17cm. The shortest side is 7cm shorter than the middle side. Find the lengths of the sides.

really stuck. Have tried loads but my answers are wrong.

2. Originally Posted by lra11
1. find the value of k for which y= kx+1 is a tangent to the curve y^2 = 8x
.
y=kx+1 is tangent to $y^2=8x$..
therefore, they have a unique common point, say (a,b), that satisfies both equation.. so $b=ka+1 \implies b^2 = k^2a^2 + 2ka + 1$ and $b^2=8a$
if we equate the $b^2$, we have $k^2a^2 + 2ka + 1 = 8a \implies k^2a^2 + 2ka - 8a + 1 = k^2a^2 + a(2k - 8) + 1 = 0$

this is a quadratic equation hence, we can solve for the value of a:
$a = \frac{8-2k \pm \sqrt{(2k-8)^2 - 4k^2}}{2k^2}$

we want a unique solution for a, so we set the discriminant be equal to 0..

$(2k-8)^2 - 4k^2 = 0$.. solve for k.. Ü

Originally Posted by lra11
2. the hypotenuse of a triangle measures 17cm. The shortest side is 7cm shorter than the middle side. Find the lengths of the sides.

really stuck. Have tried loads but my answers are wrong.
if a is the lenth of the longer side, then a-7 is the length of the shorter side,

since "hypotenuse" came, it can be assumed that this is a right triangle, hence $17^2 = a^2 + (a-7)^2$.. do the algebra..