values-find word problem

**lra11** · November 18th 2007, 01:17 PM

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.

**kalagota** · November 19th 2007, 06:36 AM

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..

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