For what value of K , y=6x+k is a tangent to curve y=7x^1/2?

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Answer 1 · 2018-05-02T14:46:53

#k=49/24#

#y=6x+k# #rarr# ①
#y=7x^(1/2)# #rarr# ②

Apply simultaneous equation, ①#=#②,

#6x+k=7x^(1/2)#

Move #7x^(1/2)# to the left,

#6x-7x^(1/2)+k=0#

Apply discriminant, #b^2-4ac=0#,

#(-7)^2-4(6)(k)=0#

Simplify,

#49-24k=0#

Add #24k# to both sides,

#49=24k#

Divide,

#k=49/24# :D

Answer 2 · 2018-05-02T15:12:00

# 49/24#.

Suppose that, the eqn. of tgt. line #t# to the curve #C:y=7x^(1/2)#

at the point #P(x_0,y_0)# is #t : y=6x+k#.

Let us note that the slope of #t# is #6#.

On the other hand, we know that, the slope of #t# at #P# is

#[dy/dx]_(P(x_0,y_0))#.

#"Now, "y=7x^(1/2) rArr dy/dx=7*1/2*x^(1/2-1)=7/(2sqrtx)#.

#:. [dy/dx]_(P(x_0,y_0))=7/(2sqrtx_0)#.

#"Clearly, "7/(2sqrtx_0)=6#.

#:. sqrtx_0=7/12#.

#"But, "P(x_0,y_0) in C. :. y_0=7sqrtx_0=7*7/12#.

#"At the same "P in t : y=6x+k#, also.

#:. y_0=6x_0+k, or, k=y_0-6x_0=7*7/12-6(7/12)^2#.

# :. k=7^2/12(1-6/12)=(49/12)(1/2)=49/24#, is the desired

value!

Enjoy Maths.!