Find the equation of tangent parallel to the secant joining the points on the curve #y=x^2# at #x=-1# and #x=2#?

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Answer 1 · 2017-03-15T07:00:03

Equation is #4x-4y-1=0# and corresponding #x#-coordinate is #1/2#

#y=x^2# is a parabola. When #x=-1#, #y=1# and when #x=2#, #y=4#.

Hence, we are seeking a tangent i.e. parallel to the secant joining points #(-1,1)# and point #(2,4)#.

As the slope of secant is #(4-1)/(2-(-1))=3/3=1#,

we are seeking a tangent with a slope of #1#.

Slope of tangent is given by #(dy)/(dx)# and as #y=x^2#, #(dy)/(dx)=2x#

Hence, for tangent we should have #2x=1# i.e. #x=1/2#

but at #x=1/2#. #y=(1/2)^2=1/4#

Hence we are seeking the tangent at #(1/2,1/4)# with slope #1#

and hence equation is #y-1/4=x-1/2# or #4x-4y-1=0#

graph{(y-x^2)(4x-4y-1)(y-x-2)=0 [-4.71, 5.29, -0.56, 4.44]}