Question

the tangent to the graph of y=sqrt x at the point P (16,4) intersects the x-axis at Q. Prove that the y-axis bisects PQ.

Answer 1

How much longer?

Answer 2

How much longer?

Answer 3

How much longer?

Answer 4

How much longer?

Answer 5

\[y= \sqrt{x}\]\[\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\]at P\[x=16\] so gradient of tangent at P
\[\frac{dy}{dx} = \frac{1}{2\sqrt{16}}=\frac{1}{8} = m\]staright line equation:\[y - y _{0} = m(x-x_{0})\]at P \[y=4 , x = 16\]\[y - 4 = \frac{1}{8}(x - 16)\]so eq of tangent at P is:\[8y - 32 = x - 16\] now finding Q, where it intersects x axis => y = 0 so:
\[-32 = x - 16\]\[x = -16\] so we now know P(16,4) and Q(-16,0)
to find where the tangent intersects y axis we set x = 0
\[8y - 32 = -16 \text{ -> }y = 2\] call this point R:
we now know P(16,4) , Q(-16,0), and R(0,2)

which the tangent passes through. now all you need to do is to show
distance PR = distance RQ