Here is the question. It states:
A circle of radius 1 unit touches positive x -axis and positive y -axis at A and B respectively. A variable line passing through origin intersects the circle in two points D and E . If the area of the triangle DEB is maximum when the slope of the line is $m$ , then the value of $\frac{1}{m^2}$ is
When I tried plotting its graph it came out to be something like this:
Now When I thought, I Felt like the triangle with maximum area should be an equilateral triangle.
So from the center of the circle I joined a line to B which makes 90 with the x axis. This line should also bisect the angle of the equilateral triangle so formed. Hence If I assume point D to be the point intersecting the circle on left. Then the line DB should have a slope $\tan(120) = -\sqrt{3}$. This line shall make and angle 60 with the line DE, so I calculated that and the slope comes out to be $m = 0$
However the answer to $\frac{1}{m^2}$ is $3$ and I am not able to understand how. Any hint/suggestions are appreciated. Thanks for your time
Best Answer
Let $D = (x_1,y_1)$ and $E = (x_2,y_2)$. Then, $$\text{ar}(\triangle DEB) = \frac{1}{2} \times \text{base } \times \text{height}\\ \text{ar}(\triangle DEB) = \frac{1}{2}(DE)(BH)$$ where $H$ is the foot of perpendicular from $B$ onto line $DE$. Suppose the equation of $DE$ is given by $y = mx$. We want to find $m$, such that the area of $\triangle DEB\ $is maximum. We know that $B = (0,1)$ and $$BH = \frac{1}{\sqrt{1+m^2}}$$ We want to find $DE$, and then maximize $\text{ar}(\triangle DEB)$. $$DE = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = |x_1-x_2|\sqrt{1+m^2}$$ since $y_1 = mx_1$ and $y_2 = mx_2$ (we know that $D$ and $E$ lie on $y = mx$). $$\text{ar}(\triangle DEB) = \frac{1}{2}|x_1 - x_2|$$ The equation of the circle is $(x-1)^2 + (y-1)^2 = 1$. Substituting $y = mx$ in this, we obtain a quadratic in $x$, with roots $x_1$ and $x_2$. With some work, you may be able to see that the equation we obtain is $$(m^2+1)x^2 - (2m+2)x + 1 = 0$$ I'm sure you know how to find $x_1 + x_2$ and $x_1x_2$ given this quadratic equation, using simply the coefficients. You'll get $$x_1+x_2 = \frac{2(m+1)}{m^2+1}$$ $$x_1x_2 = \frac{1}{m^2+1}$$ $$|x_1 - x_2| = \sqrt{(x_1+x_2)^2 - 4x_1x_2} = \frac{\sqrt{8m}}{m^2+1}$$ Finally, you'll get $$\text{ar}(\triangle DEB) = \frac{\sqrt{2m}}{m^2+1} = \frac{\sqrt{2}}{m^{3/2} + m^{-1/2}}$$ To maximize $\text{ar}(\triangle DEB)$, you want to minimize the denominator, i.e. $m^{3/2} + m^{-1/2}$ subject to $0 < m < \infty$ (otherwise the line $y = mx$ will not intersect the circle). Differentiating with respect to $m$ will yield $$m = \frac{1}{\sqrt 3}$$ Another way to look at this would be to plot $$f(x) = \frac{\sqrt 2}{x^{3/2} + x^{-1/2}}$$ which attains maxima at $\displaystyle x = \frac{1}{\sqrt{3}} \approx 0.577$.
Hope that helps!
My solution is more analytic, less geometric, but I would be interested to see more geometric approaches to this problem in other answers!