Differentiation Show that if the tangent to y=ekx at (a, eka) passes  through the origin then a=1/k

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1. Show that if the tangent to y=ekx at (a, eka) passes through the origin then a=1/k

Solution:

Differentiating the given function, we get y'=kekx. At the point (a, eka) the slope of the tangent is y'=ke^ka . The equation to the tangent is y-e^ka= ke^ka (x-a). If the tangent passes through origin, (0, 0) has to lie on the line. Hence we have 0-eka= keka (0-a).
The above equation gives -1=k(-a)1=ak a=1/k.

2. Find the value of a and b so that the line 2x +3y = a is tangent to the graph of f(x)=bx2 at the point where x = 3.
Solution:

We have f(x)=bx2. Differentiating, we get f'(x)=2bx. When x=3, we get f'(3)=6b. Hence the equation to the tangent at the point (3, 9b) is given as y-9b=6b(x-3).
Simplifying, we get the equation to the tangent as 6bx-y=9b. But 2x+3y=a is given as the tangent, hence we should have . First two ratios give, b=-1/9. Last two ratios give a=-27b=3.