Solve Dxdy Word Problem: Height of Bulb at $(1.25, 0)$

[karush]

View attachment 4671

how high h must the bulb in the figure be if point $(1.25,0)$
is right on the edge of the illuminated region.
${x}^{2}+{y}^{2}=1$

$y' = -x/y$

$\frac{-h}{13/4}=\frac{-y}{5/4-x}=-\frac{x}{y}$

hopefully if the is setup ok what is next?

 

[Greg]

The equation for the circle is incorrect.

 

[karush]

I copied the figure wrong so OP has pic from book

 

[Evgeny.Makarov]

This can be solved using geometry. Let $O$ be the origin, $C$ be the point of tangency of the sloped line and the circle, and let $A$ have coordinates $(1.25,0)$. Then the right triangle $OCA$ is similar to the large right triangle with side $h$. From there $h$ can be found.

What do you mean by dxdy in the thread title?

 

[karush]

$-y^2 = -5/4 x + x^2$
$5/4 x = x^2 + y^2 = 1$
$x = 4/5$
$y = 3/5$
$y-3/5 = -4/3 (x-4/5)$
$4x+3 y=5$
When $x=-2$ Then $y=h=13/3$

 

[MarkFL]

I would likely work this as follows:

The equation of the line is given by:

$$y=\frac{h-0}{-2-\frac{5}{4}}\left(x-\frac{5}{4}\right)=-\frac{h}{13}(4x-5)$$

Now, plug this into the equation of the circle:

$$x^2+\left(-\frac{h}{13}(4x-5)\right)^2=1$$

$$169x^2+h^2\left(16x^2-40x+25\right)=169$$

$$\left(16h^2+169\right)x^2-40h^2x+\left(25h^2-169\right)=0$$

Require the discriminant to be zero since the line and circle are tangent:

$$\left(-40h^2\right)^2-4\left(16h^2+169\right)\left(25h^2-169\right)=0$$

This reduces to:

$$(3h-13)(3h+13)=0$$

Discarding the negative root, we are left with:

$$h=\frac{13}{3}$$

 

[karush]

Well that was very helpful.

It was hard to see how to use the circle in this.

 

[Evgeny.Makarov]

We can use still another general fact that a tangent to a circle $x^2+y^2=r^2$ at point $(x_0,y_0)$ is
\[
xx_0+yy_0=r^2.\qquad(*)
\]
(More generally, the tangent to $(x-x_c)^2+(y-y_c)^2=r^2$ at point $(x_0,y_0)$ is $(x-x_c)(x_0-x_c)+(y-y_c)(y_0-y_c)=r^2$.) Substituting $x=5/4$ and $y=0$ into (*) we find $x_0=4/5$ and from $x^2+y^2=1$ it follows that $y_0=3/5$. Thus, the tangent equation is $\frac{4}{5}x+\frac{3}{5}y=1$, from where $y=13/3$ when $x=-2$.