How do you find the equation of the tangent and normal line to the curve #y=sqrtx# at x=9?

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Answer 1 · 2017-01-06T12:08:27

eqn. tgt: #" "x-6y+27=0#

eqn. normal #" "6x+y-57=0#

To find both the equation of the tangent and normal we will use

#y-y_1=m(x-x_1)#

where #m=#gradient; #(x_1,y_1)=# the coordinate in question

just considering the # +sqrt#

#y=sqrtx, y_1=sqrt9=3#

#(x_1,y_1)=(9,3)#

the problem now is to find the gradients

tangent

#m_t=(dy)/(dx)=d/(dx)(x^(1/2))=1/2x^(-1/2)#

#m_t(9)=((dy)/(dx))_(x=9)=1/2xx1/sqrt9=1/6#

eq tgt

#y-3=1/6(x-9)#

#6y-18=x-9#

#6y-x-27=0#

#x-6y+27=0#

normal

the normal is perpendicular to the tangent, by definition. So we can use the fact that the product of the gradients of perpendicular lines is #(-1)#

#m_txxm_n=-1#

#1/6xxm_n=-1#

#m_n=-6#

eqn normal

#y-3=-6(x-9)#

#y-3=-6x+54#

#6x+y-57=0#