If $(u,v)$ is a point on $4x^2+a^2y^2=4a^2$,where $4

Question

If $(u,v)$ is a point on $4x^2+a^2y^2=4a^2$,where $4<a^2<8$,that is farthest from $(0,-2)$ then $u+v$ is equal to?

My Approach:

I took a parametric point $(t,4-4t^2/a^2)$.And then tried to find the minima of the distance.But that is too lengthy method.Any other suggestions?

Answer 1

Use the method of Lagrange multipliers. In your problem you will get a system of 3 equations in 3 unknowns:
$2x=8xt$

$2y+4=2a^2yt$

$4x^2+a^2y^2=4a^2$

Solving you get

$t = 1/4$

x = $a^2\sqrt(a^2-8)\frac{1}{a^2-4}$

y = $\frac{8}{a^2-4}$

Answer 2

HINT:

WLOG any point on the ellipse $$(a\cos t,2\sin t)$$

So, if the distance is $d,$

$$d^2=(a\cos t)^2+(2+2\sin t)^2=8+4\sin t+(a^2-4)\cos^2t$$ $$=8+\sqrt{4^2+(a^2-4)^2}\sin\left(t+\arccos\dfrac4{\sqrt{4^2+(a^2-4)^2}}\right)$$

$$\le8+\sqrt{4^2+(a^2-4)^2}$$

Now $4<a^2<8,0<a^2-4<4\implies(a^2-4)^2\le4^2$

So, the maximum value i.e., the equality occurs if $t+\arccos\dfrac4{\sqrt{4^2+(a^2-4)^2}}=2m\pi+\dfrac\pi2\iff t=2m\pi+\arcsin\dfrac4{\sqrt{4^2+(a^2-4)^2}}$

and $a^2-8$ is maximum which can be made arbitrarily small but will remain negative.

Answer 3

Let $A(0,-2)$.

If $P(x,y)$ is a point on $4x^2+a^2y^2=4a^2$, then $$\begin{align}AP^2&=x^2+(y+2)^2 \\\\&=a^2-\frac{a^2y^2}{4}+(y+2)^2 \\\\&=\underbrace{\frac{4-a^2}{4}}_{\text{negative}}\bigg(y-\frac{8}{a^2-4}\bigg)^2+\frac{a^4}{a^2-4}\end{align}$$

Let $f(y):=\frac{4-a^2}{4}\bigg(y-\frac{8}{a^2-4}\bigg)^2+\frac{a^4}{a^2-4}$.

Then, $Y=f(y)$ is a downward parabola with $(\text{axis of symmetry})=\frac{8}{a^2-4}\gt 2$.

Since $y\le 2$, considering the graph of $Y=f(y)$, we get $$AP^2=f(y)\le f(2)=16$$ The equality is attained only when $(x,y)=(0,2)$.

Therefore, we get $(u,v)=(0,2)$, and so $$\color{red}{u+v=2}$$

Answer 4

Provided $4x^2+a^2y^2=4a^2$, we have to maximize $$x^2+(y+2)^2=a^2-a^2y^2/4+(y+2)^2=f(y).$$ Since the range of $y$ is $[-2,2]$ and $$f'(y)=(2-a^2/2)y+4\ge$$ $$(2-a^2/2)\cdot 2+4=8-a^2>0,$$ the function $f$ is increasing on the provided range and so $f$ attains its maximum at $y=2$. Then $x=0$ and so $u+v=x+y=2$.

Answer 5

Here is yet an other solution, that tries to work geometrically as much as possible. If computations are started, then only for a specific geometric issue, and they should be of low algebraic level. Well, i am also needing a continuity argument.

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The given constraint connecting $x,y$, $4x^2+a^2y^2=4a^2y^2$, is the ellipse of equation (divide by the R.H.S.) $$ (E)\ :\qquad \left(\frac xa\right)^2 +\left(\frac y2\right)^2 = 1\ . $$ We search a/the point $(u,v)$ on $(E)$ which is at maximal distance from the given point $P(0, -2)$. It turns out that this point is $Q(0, 2)$. Let us show this. The distance between $P$ and $Q$ is $PQ=4$. Is it the maximal distance? Before we start to answer the question, it may be useful to "see" how the problem is "working". Here is a special case with $a$ almost equal to $3$, and it is left open if the condition $a\le 2\sqrt 2$ is satisfied:

In red we have the ellipse $(E)$, and its point $P=(0,-2)$. Using radius four, we draw the circle centered in $P$ with radius $PQ=4$. Both the circle and the ellipse are tangent in $Q$ to the horizontal line $y=2$ through $Q$. How is the problem moving, when sliding the position of $(\pm a,0)$? We expect for $a=2$ that the ellipse becomes a circle centered in the origin, thus completely "inside" the other circle, then making $a$ bigger and bigger the ellipse will have the "upper" part getting closer and closer to the horizontal line $y=2$ through $Q$, so at some point it may also go "beyond" the limits of the disk... When does this happen? So we have a geometrical question.

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Is the ellipse completely inside the (closed interior disk of) the circle?

If yes, then the maximal value is taken in $Q=(0,2)$, the sum of the components of $Q$ is $0+2=2$ (and has no sense in the given geometric, or optimization context, but is a way of published examination problems to give the answer without giving it), and we are done.

If not, reductio ad absurdum, let us assume this, then consider the difference function of the two continuous functions on $[-a, a]$ determined by the two graphs in the upper half plane, the one for the circle, minus the one for the ellipse. By assumption, the difference is $0$ in $0$, takes also positive values, and in $\pm a$ we have a negative value. So the difference also takes a further zero value in $(0,a)$ and/or $(-a,0)$ (by symmetry). So the circle and the ellipse intersect in a further point. Let us intersect the two curves, see if this is possible: $$ \left\{ \begin{aligned} x^2 + (y+2)^2 &= 4^2\ ,\\ x^2 +\frac 14 a^2y ^2 &= a^2\ . \end{aligned} \right. $$ We eliminate $x$ by subtracting the equations, we obtain one equation in $y$ with parameter $a$, which is in its simpler form: $$ (a^2-4)y^2 - 16y + 4(12-a^2)=0\ . $$ The case $a=2$ is easily considered geometrically, $(E)$ is a circle, and it is contained in the (closed) interior disk of the other circle. So assume $a>2$, making the above equation an equation of degree two. One solution is $y=2$ (since it was also in the initial equation), and the other one is (Vieta for the product of the roots) such that $y+2=\frac {16}{a^2-4}$. We insert in the first equation to get the value of $x$, which satisfies: $$ x^2 = 4^2 -(y+2)^2 = 4^2 -\frac{16^2}{(a^2-4)^2} =\frac{4^2}{(a^2-4)^2}((a^2-4)^2-4^2) =\frac{4^2}{(a^2-4)^2}a^2(a^2-8)\le 0\ . $$ In the only case where there is a real $x$ solution, we have $a^2-8=0$, thus $x=0$, and the corresponding point is again $Q$. Contradiction. Our assumption is false. This leads to the wanted conclusion.

$\square$

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Note: The above computation explains also why in case of $a^2-8\ge 0$ we have a further intersection of the two conics, and the maximal distance is taken elsewhere. We have to draw further concentric circles centered in $P=(-2,0)$, and see which one in the family is touching the ellipse.