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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
 #1
avatar+118069 
+1

(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

 

 

cool cool cool

 Mar 1, 2021
 #2
avatar+118069 
+1

(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

 

 

cool cool cool

 Mar 1, 2021

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