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Tangents are drawn to $x^2+y^2=16$ from the point P $(0,h)$. These tangent stuff meet the x-axis at $A$ And $B$. Find h if area of Triangle $PAB$ is minimum.

The answer of h is $sqrt.(32)$ which shows that this point lies on director circle ⭕️. How do I get a proof that intuitively proves this?

If I assume x-intercept of both tangents to be $(a,0) and (-a,0)$ Then area of Triangle is $ah$. Which is in the form of a rectangular hyperbola where I am guessing the minimum value of $ah$ will be when a=h

Ashish Raj Shukla
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3 Answers3

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A calculation of $h$ below:

Look at $\triangle OBP$.(Symmetry about $y-$axis)

Call point of tangency $T$.

$|OT| =r$.

Consider right $\triangle OTP:$

$|PT|^2 = h^2 -r^2 , |PT| =\sqrt{h^2-r^2}.$

For right $\triangle OBP:$

$|PT||TB| = r^2.$

Area of $\triangle OBP:$

Area $= (1/2)(|PT| +|TB|)r=$

$(1/2)(|PT| + r^2/(|PT|)) .$

AM-GM:

$(|PT| +r^2/|PT|) \ge $

$2 \sqrt{(|PT|)(r^2/|PT|)} = 2r$.

Hence : Area $\ge r $.

Equality for $|PT| = r^2/|PT|$, or

$|PT|= r= 4.$

$h^2= |PT|^2+r^2= 2r^2 = 32$

$h=\sqrt{32}.$

Hence?

Peter Szilas
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A bit of trigonometry:

Note the symmetry about $y-$axis:

Look at $\triangle OBP$ (right half of the original).

Call point of tangency $T$.

$\angle BOT = \alpha.$

Then :

$\tan \alpha = |BT| /r$

Area $\triangle OBT =(1/2)|BT|r=$

$(1/2)r^2\tan \alpha$.

Likewise :

Area $\triangle OTP = (1/2)|TP|r=$

$(1/2)r^2\tan (π/2-\alpha)$.

$S:=$ Area $\triangle OBP =$

$(1/2)r^2(\tan \alpha + \tan (π/2-\alpha)).$

Need to minimize $S$:

$S:= \tan \alpha + \tan(π/2- \alpha)$, $0 < \alpha <π/2$.

$z: =\tan \alpha, z>0$.

$S= z +1/z$.

AM GM:

$S \ge 2\sqrt{z(1/z)} =2$.

Equality for $z=1/z$, i.e $z=1$.

Hence

$ \tan \alpha =1$, i.e. $\alpha =π/4$, and

$\angle OPT =π/4.$

Peter Szilas
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You have a semicircle above the $x$-axis defined by: $$f(x):=\sqrt{16-x^2}$$

Let $O=(0,0)$ be the center of the semicircle $f$. To draw a line tangent to $f$, connect find the midpoint of $OP$, which is $M_{OP}=\left(0,\frac h2\right)$. Now, draw a circle centered at $M_{OP}$ with $r=h/2$. Now we defined the function: $$g(x):=\frac{1}{2} \left(\sqrt{h^2-4 x^2}+h\right)$$ Now the points where $f(x)=g(x)$ are the points of tangency, for which we have: $$\left(-\frac{4 \sqrt{h^2-16}}{h},\frac{16}{h}\right),\left(\frac{4 \sqrt{h^2-16}}{h},\frac{16}{h}\right)$$

Now, write the functions for the lines tangent to $f$: $$h_1(x):=\frac{1}{4} \sqrt{h^2-16} x+h\\ h_2(x):=-\frac{1}{4} \sqrt{h^2-16} x+h$$

Now $h_1$ and $h_2$ intersect the $x$-axis (i.e. their zeroes) at: $$h_1:\left\{x\to -\frac{4 h}{\sqrt{h^2-16}}\right\}\\ h_2:\left\{x\to \frac{4 h}{\sqrt{h^2-16}}\right\}$$ Solving for the distance between these two points (i.e. the length of the base of the triangle), we get: $$AB=\frac{8 h}{\sqrt{h^2-16}}$$ The height of $\triangle PAB$ is $h$, therefore the area of $\triangle PAB$ is: $$A_{\triangle PAB}=\frac12\cdot AB\cdot h=\frac12 \cdot \frac{8 h}{\sqrt{h^2-16}} \cdot h = \frac{4 h^2}{\sqrt{h^2-16}}$$

To minimize $A_{\triangle PAB}$, simply take the derivative and set it to $0$: $$\left(\frac{4 h^2}{\sqrt{h^2-16}}\right)'=0\\ \frac{4 h \left(h^2-32\right)}{\left(h^2-16\right)^{3/2}}=0\\ 4 h ( h^2-32 )=0$$

Therefore, you have the following solutions: $$\left\{\{h\to 0\},\left\{h\to -4 \sqrt{2}\right\},\left\{h\to 4 \sqrt{2}\right\}\right\}$$

We know that $\angle A=\angle B$ because $PAB$ is isosceles. Then angle $A$ is $\arctan m_1$, where $m_1$ is the slope of the line passing through $P$ and $A$. Therefore: $$\angle A=\arctan \left(\frac{\sqrt{h^2-16}}{4}\right)$$ Since we have $h=\pm4\sqrt2$ as a solution, then: $$\angle A=\arctan-1=\frac\pi4 \iff \bbox[10px, border:2px black solid]{\therefore P=\frac\pi2}$$

John Glenn
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