Show that the curves whose equations are y= #sqrt(3x+1)# and y=#sqrt(5x-x^2)# have the same gradient at their point of intersection. Fina the equation of the common tangent at this point?

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Can you please show the working out.

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Answer 1 · 2017-10-18T03:41:33

The equation of the common tangent is #y = 3/4x + 5/4#

The obvious first step would be to find their points of intersection.

#sqrt(3x + 1) = sqrt(5x - x^2)#

Squaring both sides, we get:

#3x + 1 = 5x -x^2#

#x^2 - 2x + 1 = 0#

#(x - 1)(x - 1) = 0#

#x = 1#

This point will be #y = sqrt(3(1) + 1) = 2#. Hence, the point of tangency we are required to use is #(1, 2)#.

We now find the derivative of both functions. We can use the chain rule for this.

#d/dx(sqrt(3x+ 1)) = 3/(2sqrt(3x +1))#

#d/dx(sqrt(5x -x^2)) = (5 - 2x)/(2sqrt(5x - x^2))#

The gradient of the curve at that point is given by evaluating #x = a# into the derivative.

#m_ 1 = 3/(2sqrt(3(1) + 1)) = 3/4#

#m_2 = (5 - 2(1))/(2sqrt(5 - 1)) = 3/4#

The numbers are equal, hence the gradients are equal and the tangent line will be identical.

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

#y - 2 = 3/4(x - 1)#

#y - 2 = 3/4x- 3/4#

#y = 3/4x + 5/4#

Hopefully this helps!