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The tangent, $ℓ,$ to the curve

y = \sqrt{x}

at the point $P$ has a $y$ -intercept of $1.$

For (a), let

P = (a, \sqrt{a})

Just pretend you know what $a$ is and find the equation of the tangent at $P$ as usual.

You have

P = (a, \sqrt{a})

Now,

y = \sqrt{x} = x^{\frac{1}{2}}

\frac{d y}{d x} = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}

That means the tangent at $P$ (i.e. where $x = a$ ) has a gradient of

m = \frac{1}{2 \sqrt{a}}

Now use

y - y_{0} = m (x - x_{0})

You know that $(0, 1)$ lies on this line so you can find $a .$

For (b), draw a picture!

You will need to calculate the equation of the tangent line, $ℓ .$