Find the point on ellipse $4x^2+y^2=4$ that is furthest from $(1,0)$

Question

The equation of an ellipse is $4x^2 + y^2 = 4$. Find the furthest point $(x,y)$ from $(1,0)$.

Additional information from comment: First, I transformed this equation into the general formula and the result I got was: $$\frac {x^2}1 + \frac{y^2}{2^2} = 1.$$

My professor used this formula to find the distance between the 2 points: $$\sqrt{(x-1)^2 + (y-0)^2}$$ but I don't understand where this formula comes from.

Answer 1

Let $d$ be the distance from the point $(1,0)$ to a point on the ellipse $(x,y)$, then $d^2 = (x-1)^2+y^2= (x-1)^2 + 4-4x^2= x^2-2x+1+4-4x^2 = -3x^2-2x+5 $. This is a quadratic function and you know it attains a max at $x = -\frac{b}{2a} = - \frac{-2}{2(-3)} = -\frac{1}{3}$. Thus $d^2_{\text{max}} = -3\left(\frac{-1}{3}\right)^2-2\left(\frac{-1}{3}\right)+5 = \frac{16}{3}$ . So $d_{\text{max}} = \dfrac{4}{\sqrt{3}}$. Hope this helps !

Answer 2

The diagram below depicts the ellipse, and the right triangle involved in the distance formula:

In this diagram, the point on the ellipse has coordinates $(x, y)$ in general. (Here, I've chosen $x = -\frac12$ and $y = \sqrt{3}$, but it could be any point on the ellipse.) The two legs are represented by the blue line segment, which has length $|x-1|$, and the green line segment, which has length $|y|$. The hypotenuse—the orange line segment—is the desired distance $r$.

From Pythagoras, we have

$$ r^2 = (|x-1|)^2+(|y|)^2 = (x-1)^2+y^2 $$

where we can get rid of the absolute value because any real value $z$ and its absolute value $|z|$ have the same square. Take the square root of both sides (since $r \geq 0$), and we get

$$ r = \sqrt{(x-1)^2+y^2} $$

Answer 3

The distance between 2 points (say $P(x_1, y_1)$ and $Q(x_2, y_2)$) in coordinate geometry is given by the following equation: $d= \sqrt{(x_1 - x_2)^2+(y_1 - y_2)^2}$. This equation is derived by constructing a right angle triangle between the points $P$ and $Q$, and using Pythagorus's theorem to find the length (or distance between the 2 points) of the hypotenuse. (A link to an example of this sort)

So I believe your professor has said we are trying to find the distance between a point $(1,0)$ and it's furtherest point, $(x,y)$ (where $(x,y)$ is on the ellipse). Thus, the distance between the 2 points is $d= \sqrt{(x - 1)^2+(y - 0)^2}$.

Answer 4

WLOG $(x,y)$ can be chosen to be $(\cos t,2\sin t)$

If $d$ is the distance between $(x,y);(1,0)$

$d^2=\cos^2t-2\cos t+1+4\sin^2t=5-2\cos t-3\cos^2t$

$=5+\dfrac13-3\left(\cos t+\dfrac13\right)^2\le5+\dfrac13$