Two Dimensional Analytical Geometry-II — Flashcards
Tamil Nadu Board · Class 12 · Mathematics
30 flashcards for Two Dimensional Analytical Geometry-II (Tamil Nadu Board Class 12 Mathematics) to test yourself on key terms and facts.
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Find the general equation of a circle with centre (-3, -4) and radius 3 units.
Answer
Step 1: Use standard form (x - h)² + (y - k)² = r² Step 2: Substitute h = -3, k = -4, r = 3 (x - (-3))² + (y - (-4))² = 3² Step 3: Expand (x + 3)² + (y + 4)² = 9 Step 4: Expand further x² + 6x + 9 + y…
Find the centre and radius of the circle x² + y² - 6x + 4y + c = 0 for all possible values of c.
Answer
Step 1: Use the general form x² + y² + 2gx + 2fy + c = 0 Step 2: Compare coefficients: 2g = -6 → g = -3, 2f = 4 → f = -2 Step 3: Centre = (-g, -f) = (-(-3), -(-2)) = (3, 2) Step 4: Radius = √(g² + f² …
Determine whether the point (2, 3) lies inside, on, or outside the circle x² + y² - 6x - 8y + 12 = 0.
Answer
Step 1: Use the position test formula for point (x₁, y₁) Step 2: Substitute (x₁, y₁) = (2, 3) into x² + y² - 6x - 8y + 12 Step 3: (2)² + (3)² - 6(2) - 8(3) + 12 Step 4: = 4 + 9 - 12 - 24 + 12 = -11 St…
Find the equation of the circle whose diameter is the line segment joining the points (-4, -2) and (1, 1).
Answer
Step 1: Use diameter form (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0 Step 2: Substitute (x₁, y₁) = (-4, -2) and (x₂, y₂) = (1, 1) Step 3: (x - (-4))(x - 1) + (y - (-2))(y - 1) = 0 Step 4: (x + 4)(x - 1) …
Find the equations of the tangent and normal to the circle x² + y² = 25 at point P(-3, 4).
Answer
Tangent at (x₁, y₁) to circle x² + y² = a² is xx₁ + yy₁ = a² Step 1 (Tangent): Substitute (x₁, y₁) = (-3, 4) and a² = 25 x(-3) + y(4) = 25 Step 2: -3x + 4y = 25 Answer (Tangent): -3x + 4y = 25 or 3x…
If y = 4x + c is a tangent to the circle x² + y² = 9, find the value of c.
Answer
Step 1: Use tangency condition for line y = mx + c to circle x² + y² = a² Condition: c² = a²(1 + m²) Step 2: Here m = 4, a² = 9 Step 3: c² = 9(1 + 4²) = 9(1 + 16) = 9(17) = 153 Step 4: c = ±√153 = ±3√…
Find the equation of the parabola with focus (-√2, 0) and directrix x = √2.
Answer
Step 1: Identify vertex - midpoint between focus and directrix Vertex x-coordinate = [(-√2 + √2)/2] = 0 Step 2: Parabola opens to the LEFT (focus is left of directrix) Step 3: Distance from vertex to …
Find the vertex, focus, and equation of directrix for the parabola x² - 4x - 5y - 1 = 0.
Answer
Step 1: Rearrange x² - 4x = 5y + 1 Step 2: Complete the square x² - 4x + 4 = 5y + 1 + 4 Step 3: (x - 2)² = 5(y + 1) Step 4: This is in form (x - h)² = 4a(y - k) where h = 2, k = -1 Step 5: 4a = 5 → a …
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