Minimum value of slope of line tangent to ellipse

Question

I would like to solve the following problem.

Given that the point $p$ lies on the ellipse $$\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1$$ and the line $y = ax$ passes through $p$, let $n$ be the smallest positive value of $a$. Compute $n$.

What first came to my mind was finding the slope of the line tangent to the ellipse. This would require me to set the discriminant of the resulting quadratic in $x$ to be set to $0$.

However, the numbers in the denominator make the calculations a bit annoying. Is there a better way to solve the problem? Thanks!

Answer 1

Not the best in math way answer because it is not general solution but it is based on my comment:

My idea is (for easy solution without hard work) to find the line $y=αx$ that is tangent in the circle that contains the ellipse and then test next few values with the real ellipse. [I know it is a hack but also know that the numbers seems to be a little bit silly for this question... Sure the solution is close to $1$ if not $1$]

The circle that contains the ellipse (don't know English name) is:

$x^2+(y-12)^2=10^2$

The tangent on this circle that can pass through $(0,0)$ is a line of function $y=\alpha \cdot x$ that has distance d=10 from the point $(0,12)$.

line is:

$\alpha \cdot x -y=0$ and point (0,12) have a distance:

$d=\dfrac{|\alpha \cdot 0 -12|}{\sqrt{\alpha^2+1}}=\dfrac{12}{\sqrt{\alpha^2+1}}=10$

Thus:

$100\alpha^2+100=144\Longrightarrow \alpha^2=\frac{44}{100}$.

We are looking for integer values of $\alpha$, and so we will start from value $1$ and test a few integers with the real ellipse [Possibly $1$ could already passes through].

Test value $\alpha=1$

$\begin{cases}y=x\\ \dfrac{x^2}{100}+\dfrac{(x-12)^2}{75}=1 \end{cases}\hspace{10pt} \begin{cases}y=x\\ x^2+\dfrac{4}{3}(x-12)^2=100 \end{cases}\hspace{10pt} \begin{cases}y=x\\ x^2+\dfrac{4}{3}x^2 +\dfrac{4}{3}144-\dfrac{4}{3}\cdot 24 \cdot x=100 \end{cases}$

$\begin{cases}\dfrac{7}{3}x^2-32\cdot x+16\cdot 12-100=0\\\end{cases}$

The last one becomes:

$7x^2- 96 x +(192-100)\cdot 3=0$

$7x^2-96 x+276=0$ with

$\Delta=(-96)^2-4\cdot 7\cdot 276\approx 10,000-7000 >0$

Solution =1

Edit: Visualizing why I found the numbers silly:

Answer 2

By substitution you can get equations for the points of intersection between the line y=ax and the ellipse; and the equations will cook down x1 = fx1(a), y1 = fy1(a), x2 = fx2(a), y2 = fy2(a), where the functions fx1,fx2,fy1, fy2 emerge from the process of substitution. The tangent point is where a value of (a) forces x1 = x2 and y1 = y2. So, you then need to solve for (a) in fx1(a) = fx2(a) and fy1(a)=fy2(a) There will actually be two tangent points. You need to find the one that has the smaller value of a.

Alternatively, you could parametrize the ellipse into a form x = Fx(tau) and y = Fy(tau). Calculate (a) from x and y: (a) = y/x. So, Fy(tau)/Fx(tau) = (a). Take the derivative d(a)/d(tau), set equal to zero and solve for tau. This will have two solutions. Plug them back in to Fy(tau)/Fx(tau) = (a) and find the one that gives the smaller value of a.

Answer 3

WLOG $p(10\cos2t,12+5\sqrt3\sin2t)$

So, $12+5\sqrt3\sin2t=10a\cos2t$

Method$\#1:$

$$12=10a\cos2t-5\sqrt3\sin2t\le\sqrt{(10a)^2+(5\sqrt3)^2}$$

$$\iff144\le100a^2+75\iff100a^2\le69$$

Method$\#2:$

Now use this to form a quadratic equation in $\tan t$

and use the idea of $x^2+y^2=1$, find min/max of $(3x+2y)^2+(x+2y)^2$

Answer 4

The coordinates $(x,y)$ of point $p$ satisfy both the equation of the ellipse and $y=ax$, therefore:

$$ \begin{align} \frac{x^2}{100} + \frac{(ax - 12)^2}{75} = 1 \quad&\iff\quad 3 x^2 + 4(ax - 12)^2 = 300 \\ &\iff\quad (4 a^2 + 3) x^2 - 96 a x + 276 = 0 \end{align} $$

The latter is a quadratic in $x$ and the condition for it to have real roots is that the (reduced) discriminant be non-negative:

$$ 0 \le \frac{1}{4}\Delta = (48a)^2-276(4a^2+3) = 12 (100 a^2 - 69) $$

The latter gives the range $|a| \ge \frac{\sqrt{69}}{10}$, and from there the smallest positive value.

Answer 5

All of these lines lie between the tangents to the ellipse through their common point—the origin—so the slopes are also bounded by the slopes of these two tangents. Thus, the problem reduces to that of finding the tangents to the ellipse that pass through the origin. There are various ways to do this, but the simplest given the symmetry of the ellipse about the $y$-axis is basically what you’ve already outlined: Substitute $y=ax$ into the equation of the ellipse, set the discriminant to zero so that there’s only one intersection between the line and ellipse, and solve for $a$.