Conics (Ellipse): Complete the Equation to Give at least 1 point

Question

The question asks:

For which values of $a$ does the conic $4x^2+16x+5y^2-40y=a$ have at least one point? (State your answer in interval notation.)

$a\in$ ___

I was able to understand that the conic mentioned is an ellipse, and completed the square to get it into the following format:

$$4(x+2)^2+5(y-4)^2=a+96$$

I'm guessing that it wants me to give a range of values to make the equation valid, but I'm not sure how to proceed from here. Any advice would be much appreciated.

Answer 1

In order for the equation to represent a non-degenerate ellipse, notice that it must be able to be rearranged into the standard form for an ellipse, which is $$\frac{\left(x - h\right)^{2}}{a^{2}} + \frac{\left(y - k\right)^{2}}{b^{2}} = 1.$$

What you have there is very close to the standard form. Notice that for the equation to be able to be translated into standard form, $c + 96 \geq 0.$ The range of $c$ is simply $\boxed{c \geq -96}.$

Answer 2

Any $c \geq -96$ is good. Infact since LHS $\geq 0$ of course $c \geq -96$ is needed.

Set $c + 96 = d \geq 0$. Then put $y= x+6$, you get

$9(x+2)^2=d$

Now check that $\Delta = \frac{4}{9}d \geq 0$.

This actually appeared as a wierd trick to me, now I get what is going on: try to intersect the ellispe with a line passing trought the center ie $(-2,4)$, this line must intersect the ellipse if it does exist!! In this case I used $y = x+6$ becuase it simplifies the equation, but you can chose any such kind of line.

Answer 3

Hint: Since $$A^2 \geq 0$$ holds for any real number $A$, a suitable lower bound for $A^2 + 5B^2$ is $$A^2 + 5B^2 \geq 0.$$

Now set $A = 2(x + 2)$ and $B = y - 4$.

Can you take it from here?