Conic Sections — Flashcards
CBSE · Class 11 · Mathematics
56 flashcards for Conic Sections (CBSE Class 11 Mathematics) to test yourself on key terms and facts. Sample: "Find the equation of a circle with centre"
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Find the equation of a circle with centre (0, 0) and radius 7.
Answer
Step 1: Use the circle equation (x - h)^2 + (y - k)^2 = r^2. Step 2: Here, h = 0, k = 0, and r = 7. Step 3: Substitute: x^2 + y^2 = 7^2. Answer: x^2 + y^2 = 49.
केंद्र (0, 0) और त्रिज्या 7 वाले एक वृत्त का समीकरण ज्ञात कीजिए।
Answer
चरण 1: वृत्त का समीकरण (x - h)^2 + (y - k)^2 = r^2 का उपयोग कीजिए। चरण 2: यहाँ, h = 0, k = 0, और r = 7 है। चरण 3: रखने पर: x^2 + y^2 = 7^2. उत्तर: x^2 + y^2 = 49.
Find the equation of a circle with centre (-3, 2) and radius 4.
Answer
Step 1: Use (x - h)^2 + (y - k)^2 = r^2. Step 2: Substitute h = -3, k = 2, r = 4. Step 3: (x + 3)^2 + (y - 2)^2 = 16. Answer: (x + 3)^2 + (y - 2)^2 = 16.
केंद्र (-3, 2) और त्रिज्या 4 वाले एक वृत्त का समीकरण ज्ञात कीजिए।
Answer
चरण 1: (x - h)^2 + (y - k)^2 = r^2 का उपयोग कीजिए। चरण 2: h = -3, k = 2, r = 4 रखें। चरण 3: (x + 3)^2 + (y - 2)^2 = 16. उत्तर: (x + 3)^2 + (y - 2)^2 = 16.
Find the centre and radius of the circle x^2 + y^2 + 8x + 10y - 8 = 0.
Answer
Step 1: Group x terms and y terms: (x^2 + 8x) + (y^2 + 10y) = 8. Step 2: Complete squares: (x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25. Step 3: Write as (x + 4)^2 + (y + 5)^2 = 49. Step 4: Compar…
वृत्त x^2 + y^2 + 8x + 10y - 8 = 0 का केंद्र और त्रिज्या ज्ञात कीजिए।
Answer
चरण 1: x पदों और y पदों को समूहित कीजिए: (x^2 + 8x) + (y^2 + 10y) = 8. चरण 2: वर्ग पूर्ण कीजिए: (x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25. चरण 3: इसे (x + 4)^2 + (y + 5)^2 = 49 के रूप में लिखिए…
A circle passes through (2, -2) and (3, 4), and its centre lies on x + y = 2. Find its equation.
Answer
Step 1: Let the centre be (h, k) and radius be r. Step 2: Since the circle passes through the points: (2 - h)^2 + (-2 - k)^2 = r^2 (3 - h)^2 + (4 - k)^2 = r^2 Step 3: Also, h + k = 2. Step 4: Solving …
एक वृत्त (2, -2) और (3, 4) से होकर जाता है, और उसका केंद्र x + y = 2 पर स्थित है। उसका समीकरण ज्ञात कीजिए।
Answer
चरण 1: केंद्र को (h, k) तथा त्रिज्या को r मानिए। चरण 2: क्योंकि वृत्त दिए गए बिंदुओं से होकर जाता है: (2 - h)^2 + (-2 - k)^2 = r^2 (3 - h)^2 + (4 - k)^2 = r^2 चरण 3: साथ ही, h + k = 2. चरण 4: इन्हें ह…
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