في القطع الناقص يتقاطع المحوران الاكبر والاصغر في

الرأسان المرافقان

ما هو المركز؟

نقطة منتصف المحور الأكبر

نهايتا المحور الاكبر

نهايتا المحور الاصغر

القطع الناقص

Authored by jory math

Mathematics

12th Grade

Used 9+ times

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12 questions

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

مركز القطع الناقص
$\frac{\left(x+3\right)^2}{36}+\frac{\left(y+1\right)^2}{9}=1$

(-3,-1)

(3,1)

(-1,3)

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

معادلة المحور الأكبر للقطع الناقص
$\frac{\left(x-3\right)^2}{4}+\frac{\left(y+2\right)^2}{16}=1$

x = -2

x = 3

y = 3

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

اتجاه القطع الناقص
$\frac{\left(x+4\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=1$
:هو

افقي

رأسي

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

طول المحور الأكبر للقطع الناقص
$\frac{\left(x-3\right)^2}{36}+\frac{\left(y+1\right)^2}{9}=1$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

معادلة الدائرة التي مركزها نقطة الأصل وقطرها 14 هي

$x+y=49$

$x^2-y^2=49$

$x^2+y^2=49$

$x^2+y^2=7$

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القطع الناقص

مركز القطع الناقص (x+3)236+(y+1)29=1\frac{\left(x+3\right)^2}{36}+\frac{\left(y+1\right)^2}{9}=136(x+3)2​+9(y+1)2​=1

معادلة المحور الأكبر للقطع الناقص (x−3)24+(y+2)216=1\frac{\left(x-3\right)^2}{4}+\frac{\left(y+2\right)^2}{16}=14(x−3)2​+16(y+2)2​=1

اتجاه القطع الناقص (x+4)29+(y+3)24=1\frac{\left(x+4\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=19(x+4)2​+4(y+3)2​=1 :هو

طول المحور الأكبر للقطع الناقص (x−3)236+(y+1)29=1\frac{\left(x-3\right)^2}{36}+\frac{\left(y+1\right)^2}{9}=136(x−3)2​+9(y+1)2​=1

في القطع الناقص يتقاطع المحوران الاكبر والاصغر في

معادلة الدائرة التي مركزها نقطة الأصل وقطرها 14 هي

ما هو المركز؟

Access all questions and much more by creating a free account

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Popular Resources on Wayground

مركز القطع الناقص
$\frac{\left(x+3\right)^2}{36}+\frac{\left(y+1\right)^2}{9}=1$

معادلة المحور الأكبر للقطع الناقص
$\frac{\left(x-3\right)^2}{4}+\frac{\left(y+2\right)^2}{16}=1$

اتجاه القطع الناقص
$\frac{\left(x+4\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=1$
:هو

طول المحور الأكبر للقطع الناقص
$\frac{\left(x-3\right)^2}{36}+\frac{\left(y+1\right)^2}{9}=1$