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# help algebra

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Find the value(s) or range of values of k for which

(a) the line y = kx - 5 is tangent to the curve x^2 = 2y + 1.

(b) the line x + 3y = k - 1 intersect the curve y^2 = 2x + 5.

Mar 1, 2021

### 2+0 Answers

#1
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(a) the line y = kx - 5 is tangent to the curve x^2 = 2y + 1

Using implicit  differentiation, the slope at  any point  on the  quadratic function (curve)  can be found as

2y = x^2  - 1

2y'  = 2x

y' =  x

Let  the point  where  the line is tangent to the  curve  be (x , x^2/2 - 1/2)

The line  will  intercept the point  (0,-5)

So  we  need  to  solve  the slope equation

( x^2/2  - 1/2  -  - 5)  / ( x - 0)  = x

(x^2/2 - 1/2 + 5)  /  x   =   x        multiply through  by x

x^2/2 + 9/2  = x^2                 multiply through by 2

x^2 + 9  = 2x^2        rearrange  as

x^2  - 9   =  0

x^2  = 9             take both roots

x = 3      or   x  = -3       =     the two values for k

The  lines  are

y = 3x  - 5      and    y   = -3x  -5

See here :   https://www.desmos.com/calculator/npleonadqb   Mar 1, 2021
#2
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(b) the line x + 3y = k - 1 intersect the curve y^2 = 2x + 5

See here  :   https://www.desmos.com/calculator/hmtcql97ih

k  =  -6

The  equation of the line is

x + 3y  =   -7   Mar 1, 2021