Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$

Question

Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$

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Let the required line is tangent to the parabola at the point $(x_1,y_1)$.It passes through $(\frac{1}{2},2)$.Its equation is $y-2=-x_1(x-\frac{1}{2})$.
This line is also secant to the the curve $y=\sqrt{4-x^2}$.
I solved $y=\sqrt{4-x^2}$ and the line $y-2=-x_1(x-\frac{1}{2})$.
I am stuck here.

Answer 1

Let $\left(t,-\frac{t^2}{2}+2\right)$ be a tangent point.

Since $\left(-\frac{x^2}{2}+2\right)'=-x$, we get an equation of the tangent line: $$y+\frac{t^2}{2}-2=-t(x-t).$$ Now, substitute $x=\frac{1}{2}$ and $y=2$, find a values of $t$ (I got $t=0$ or $t=1$) and choose a value, which you need.

Answer 2

HINT:

The equation of any straight line passing through $(1/2,2)$ is $$\dfrac{y-2}{x-1/2}=m\iff y=mx+2-m/2$$ where $2m$ is the gradient.

Let us find the intersection of this line with the given parabola.

$$mx+m-\dfrac12=2-\dfrac{x^2}2\iff x^2+2mx+2m-5=0$$ which is a quadratic equation in $x$ whose each of the two roots represents the abscissa of intersection. For tangency, both roots must be same.

Answer 3

Problem

$$ \color{red}{y = 2 - \frac{x^{2}}{2}} \tag{1} $$ $$ \color{blue}{y = \sqrt{4-x^2}} \tag{2} $$

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Equation of the line

Find the equation of the tangent line $$ y = m x + b \tag{3} $$ tangent to $\color{red}{y(x)}$

We are given the a point $$ p = \left( \frac{1}{2}, 2 \right) $$

Find the slope, $m$, and the intercept, $b$.

Slope

To be tangent to the red curve, the slope of the line must match the slope of red curve. The slope is of the red curve is $$ \color{red}{y'} = -1 $$

Intercept

The intercept is computed from $(3)$ using the point $p$: $$ b = y - m x \qquad \Rightarrow \qquad b = 1 - (-1) \frac{1}{2} = \frac{5}{4} $$

Solution

The equation for the tangent line (the dashed line below) is $$ \boxed{ y = -x + \frac{5}{4}} \tag{4} $$

Tangent point

Where does the dashed line, $(4)$, touch the red curve, $(1)$? Solve $$ \begin{align} y &= \color{red}{y} \\ % -x + \frac{5}{4} &= \color{red}{2 - \frac{x^{2}}{2}} \\ % x &= 1 \end{align} $$ Using $(4)$, we have the tangent point is $$ q = \left( 1, \frac{3}{2} \right), $$ where the dashed line touches the red curve.

Secant points

Where are the two points where the dashed line, $(4)$ intercepts the blue curve, $(2)$? Solve $$ \begin{align} y &= \color{blue}{y} \\ -x + \frac{5}{4} &= \color{blue}{\sqrt{4-x^2}} \\ \end{align} $$ The solution is $x = \frac{1}{4} \left(5 \pm \sqrt{7} \right)$. Therefore the two points define the secant chord are $$ % \frac{1}{4} \left( \left(5 - \sqrt{7}\right), \left(5 + \sqrt{7}\right) + 10 \right), \qquad % \frac{1}{4} \left( \left(5 + \sqrt{7}\right), \left(-5 - \sqrt{7}\right) + 10 \right) % $$