The length of sub-tangent at (x1,y1) to the curve y=ex5 is
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Solution
Here, y=ex5⋯(i)
and (x1,y1) lies on the curve. ⇒dydx=ex5⋅15⋯(ii)
And we know that the length of the sub-tangent =∣∣∣y1dxdy∣∣∣ =∣∣
∣
∣∣ex15⋅5ex15∣∣
∣
∣∣=5 units [using (i) and (ii)]