To determine

a) To find: the equation of the tangent to the curve y=ex that is parallel to the line x−4y=1

Answer

  The equation of the tangent to the curve y=ex that is parallel to the line x−4y=1 is given by y=14(x+ln4+1)

Explanation

Explanation:  The Slope of the tangent to the curve y=ex is equal to the slope of the line x−4y=1 as they are parallel

x−4y=1⇒y=14(x−1)

Comparing with y=mx+c, we get m=14

Since y=ex, we have y′=ex

Therefore

y′=ex=14⇒x=ln(14)=ln1−ln4=−ln4

y=ex⇒y=eln(14)=14

From Point Slope form, we have

y−y1=m(x−x1)

Therefore y−14=14(x+ln4)

 y=14(x+ln4+1)

Conclusion: y=14(x+ln4+1)

To determine

b) To find: the equation of the tangent to the curve y=ex that passes through the origin

Answer

y=ex

Explanation

Since y=ex, we have y′=ex

From Point Slope form, we have

y−y1=m(x−x1)

y−0=m(x−0)

y=mx

m=yx=exx

Also

m=y′=ex

Therefore,

ex=exx⇒x=1

m=exx=e

Hence y=mx=ex

Conclusion: The equation of the tangent to the curve y=ex that passes through the origin is y=ex