القطع الزائد

Authored by mahmoud wazer

Mathematics

12th Grade

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12 questions

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MULTIPLE CHOICE QUESTION

45 sec • 1 pt

مركز القطع الزائد
$\frac{\left(x-1\right)^2}{16}-\frac{\left(y+2\right)^2}{9}=1$

(-1,2)

(1,-2)

(-1,-2)

(1,2)

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

مركز القطع الزائد
$\frac{\left(x-1\right)^2}{16}-\frac{\left(y+2\right)^2}{9}=1$

(-1,2)

(1,-2)

(-1,-2)

(1,2)

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

معادلة المحور القاطع للقطع الزائد
$\frac{\left(y-3\right)^2}{4}-\frac{\left(x+2\right)^2}{5}=1$

y= -3

x=2

y=3

x= -2

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

معادلتا خطا التقارب للقطع الزائد
$\frac{\left(x+1\right)^2}{4}-\frac{\left(y-3\right)^2}{1}=1$

$y-3=\pm2\left(x+2\right)$

$y-3=\pm\frac{1}{2}\left(x+2\right)$

$y=\pm\frac{1}{2}x$

$y=\pm2x$

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

طول المحور القاطع في القطع الزائد
$\frac{\left(y-3\right)^2}{9}-\frac{\left(x+1\right)^2}{16}=1$

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

مركز القطع الزائد الممثل جانبا

(0,-6)

(0,6)

(6,0)

(-6,0)

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

في القطع الزائد البؤرتان
$\frac{x^2}{9}-\frac{\left(y-1\right)^2}{16}=1$

$\left(1,\pm5\right)$

$\left(\pm5,0\right)$

$\left(\pm5,1\right)$

$\left(\pm5,-1\right)$

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

في القطع الزائد الرأسان
$\frac{x^2}{9}-\frac{\left(y-1\right)^2}{16}=1$

$\left(1,\pm3\right)$

$\left(\pm3,0\right)$

$\left(\pm4,1\right)$

$\left(\pm3,1\right)$

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

في المعادلة هي h قيمة
$\frac{\left(\times+1\right)^2}{9}-\frac{\left(y+2\right)}{16}=1$

-2

-1

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القطع الزائد

مركز القطع الزائد (x−1)216−(y+2)29=1\frac{\left(x-1\right)^2}{16}-\frac{\left(y+2\right)^2}{9}=116(x−1)2​−9(y+2)2​=1

مركز القطع الزائد (x−1)216−(y+2)29=1\frac{\left(x-1\right)^2}{16}-\frac{\left(y+2\right)^2}{9}=116(x−1)2​−9(y+2)2​=1

معادلة المحور القاطع للقطع الزائد (y−3)24−(x+2)25=1\frac{\left(y-3\right)^2}{4}-\frac{\left(x+2\right)^2}{5}=14(y−3)2​−5(x+2)2​=1

معادلتا خطا التقارب للقطع الزائد (x+1)24−(y−3)21=1\frac{\left(x+1\right)^2}{4}-\frac{\left(y-3\right)^2}{1}=14(x+1)2​−1(y−3)2​=1

طول المحور القاطع في القطع الزائد (y−3)29−(x+1)216=1\frac{\left(y-3\right)^2}{9}-\frac{\left(x+1\right)^2}{16}=19(y−3)2​−16(x+1)2​=1

مركز القطع الزائد الممثل جانبا

في القطع الزائد البؤرتان x29−(y−1)216=1\frac{x^2}{9}-\frac{\left(y-1\right)^2}{16}=19x2​−16(y−1)2​=1

في القطع الزائد الرأسان x29−(y−1)216=1\frac{x^2}{9}-\frac{\left(y-1\right)^2}{16}=19x2​−16(y−1)2​=1

في المعادلة هي h قيمة (×+1)29−(y+2)16=1\frac{\left(\times+1\right)^2}{9}-\frac{\left(y+2\right)}{16}=19(×+1)2​−16(y+2)​=1

: يكون الاتجاه افقي عندما تكون

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مركز القطع الزائد
$\frac{\left(x-1\right)^2}{16}-\frac{\left(y+2\right)^2}{9}=1$

مركز القطع الزائد
$\frac{\left(x-1\right)^2}{16}-\frac{\left(y+2\right)^2}{9}=1$

معادلة المحور القاطع للقطع الزائد
$\frac{\left(y-3\right)^2}{4}-\frac{\left(x+2\right)^2}{5}=1$

معادلتا خطا التقارب للقطع الزائد
$\frac{\left(x+1\right)^2}{4}-\frac{\left(y-3\right)^2}{1}=1$

طول المحور القاطع في القطع الزائد
$\frac{\left(y-3\right)^2}{9}-\frac{\left(x+1\right)^2}{16}=1$

في القطع الزائد البؤرتان
$\frac{x^2}{9}-\frac{\left(y-1\right)^2}{16}=1$

في القطع الزائد الرأسان
$\frac{x^2}{9}-\frac{\left(y-1\right)^2}{16}=1$

في المعادلة هي h قيمة
$\frac{\left(\times+1\right)^2}{9}-\frac{\left(y+2\right)}{16}=1$