Question

please help meee :((
The system of conics has two solutions.

(x−4)2+(y+1)2=9
(x−4)29+(y+1)281=1

What are the solutions to this system of conics?

Answer 1

Answer:

• (1, -1) and (7, -1)

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Given system

• (x − 4)² + (y + 1)² = 9
• (x − 4)² / 9 + (y + 1)² / 81 = 1

Solve it by elimination

Multiply the second equation by 9 and subtract from the first one:

• (y + 1)² -  (y + 1)² / 9 = 0
• (y + 1)²/8 = 0
• (y + 1)² = 0
• y + 1 = 0
• y = - 1

Substitute y and solve for x

• (x − 4)² + (-1 + 1)² = 9
• (x - 4)² = 9
• x -4 = ± 3
• x = 4 ± 3
• x = 1 and x = 7

Answer 2

Answer:

$\left(\; \boxed{1}\:,\boxed{-1}\;\right)\; \textsf{and}\;\left(\; \boxed{7}\:,\boxed{-1}\;\right)$

Step-by-step explanation:

Given system of conics:

$\begin{cases}(x-4)^2+(y+1)^2=9\\\\\dfrac{(x-4)^2}{9}+\dfrac{(y+1)^2}{81}=1\end{aligned}$

Multiply the second equation by 81:

$\implies \dfrac{81(x-4)^2}{9}+\dfrac{81(y+1)^2}{81}=81$

$\implies 9(x-4)^2+(y+1)^2=81$

Rearrange to make (y + 1)² the subject:

$\implies (y+1)^2=81-9(x-4)^2$

Substitute into the first equation and simplify:

$\implies (x-4)^2+81-9(x-4)^2=9$

$\implies -8(x-4)^2=-72$

$\implies (x-4)^2=9$

Solve for x:

$\implies \sqrt{ (x-4)^2}=\sqrt{9}$

$\implies x-4=\pm3$

$\implies x=4\pm3$

$\implies x=1, 7$

Substitute the found values of x into the first equation and solve for y:

$\begin{aligned}x=1 \implies (1-4)^2+(y+1)^2&=9\\(-3)^2+(y+1)^2&=9\\9+(y+1)^2&=9\\(y+1)^2&=0\\y+1&=0\\y&=-1\end{aligned}$

$\begin{aligned}x=7 \implies (7-4)^2+(y+1)^2&=9\\(3)^2+(y+1)^2&=9\\9+(y+1)^2&=9\\(y+1)^2&=0\\y+1&=0\\y&=-1\end{aligned}$

Therefore, the solution to the given system of conics is:

• (1, -1) and (7, -1)