Question

exprees the standard form is 893.256​

Answer 1

Answer:

8000+900+30.2/10 5/100 6/1000

Answer 2

Answer:

$By \: completing \: the \: square, the \: \\ standard \: form \: of \: the \: hyperbola \\ is:$

$\frac{(x - 4) {}^{2} }{16} - (y - 4) {}^{2} = 1$

$with \: vertices \: at  \: (0,−4 )(0,−4)  \\ \: and \:  (8,−4)(8,−4).$

$The \: ellipse \: will \: have \: the \: same \\ center \: as \: the \: hyperbola,  \\ (4,−4)(4,−4),$

$and \: the \: foci \: are \: at \: the \: vertices \:$

$of \: the \: hyperbola, (0,−4)(0,−4) \\  and (8,−4) \: so \: we see \\ that c=4$

$These \: foci \: and \: the \: given \: point \\  (8,−10)(8,−10) form \: a \: right \\ triangle \: with \: legs \: of \: length 88  \\  (between \: the \: foci) \: and 66  \\ (from \: the \: right \: focus \: to \: the \: \\ given \: point), , and \: a \: hypotenuse \: \\ of length 10.$

$For \: an \: ellipse, \: the \: distance \: \\ from \: a \: focus \: to \: a \: point \: on \: the \\ ellipse  \: is  \: 2a=6+10=16. \: \\ Therefore, a=8 \: and \: b {}^{2} = a {}^{2} \\ - \: c {}^{2} = 64 - 16 = 48$

$So, the \: required \: ellipse \: is: \\ \frac{(x - 4) {}^{2} }{64} + \frac{(y + 4) {}^{2} }{48} = 1$

$The \: plot \: looks \: like \: this:$