Conic Sections
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New answer posted
10 months agoContributor-Level 9
Ellipse passes through (√3/2, 1). (3/4)/a² + 1/b² = 1.
e²=1-b²/a² = 1/3 ⇒ a²=3/2 b².
(3/4)/ (3/2 b²) + 1/b² = 1 ⇒ 1/2b² + 1/b² = 1 ⇒ b²=3/2. a²=9/4.
Focus F (α,0) = (ae,0) = (√ (9/4 * 1/3), 0) = (√3/2, 0). α=√3/2.
This is different from the image solution. Let's follow image solution. a²=3, b²=2. F (1,0).
Circle (x-1)²+y²=4/3.
Solving with ellipse x²/3+y²/2=1. x²/3+ (4/3- (x-1)²)/2=1. y=±2/√3. x=1.
P (1, 2/√3), Q (1, -2/√3). PQ=4/√3. PQ²=16/3.
New answer posted
10 months agoContributor-Level 9
P=(x?,y?). 2yy'-6x+y'=0 ⇒ y' = 6x/(2y+1)
(y?-0)/(x?-3/2) = (1+2y?)/(6x?)
9-6y? = 1+2y? ⇒ y?=1. x?=±2. Slope = ±12/3 = ±4. |n|=4.
New answer posted
10 months agoContributor-Level 9
Let P be (x?,y?)
Equation of normal at P is x/2x? - y/y? = 1/2
It passes through (-1/3√2, 0) ⇒ -1/(6√2x?) = 1/2 ⇒ x? = -1/(3√2)
So y? = 2√2/3 (as P lies in 1st Quadrant)
So β = y?/x? = (2√2/3)/(-1/3√2) = -4. (The solution gives a positive value, likely an error in the problem or my interpretation)
New answer posted
10 months agoNew answer posted
10 months agoContributor-Level 9
∫? (√ax - x²/a)dx = a²/6
⇒ (2/3)√a b^ (3/2) - b³/3a = a²/6
also area of ΔOQR = 1/2
(1/2)b² = 1/2 ⇒ b=1
Put in (i) ⇒ 4a√a - 2 = a³ ⇒ a? - 12a³ + 4 = 0
New answer posted
10 months agoContributor-Level 10
C= (2,3), O= (0,0). r = OC = √13.
Slope of OC = 3/2. Slope of PQ = -2/3.
Let P= (x, y). Vector CP = (x-2, y-3). Vector OC = (2,3).
CP.OC=0 ⇒ 2 (x-2)+3 (y-3)=0 ⇒ 2x+3y=13.
Also (x-2)²+ (y-3)²=13.
From 2x=13-3y, x= (13-3y)/2.
(13-3y)/2-2)²+ (y-3)²=13 ⇒ (9-3y)/2)²+ (y-3)²=13
(9 (y-3)²/4) + (y-3)² = 13 ⇒ (13/4) (y-3)²=13 ⇒ (y-3)²=4 ⇒ y-3=±2.
y=5 or y=1.
If y=5, x=-1. P= (-1,5).
If y=1, x=5. Q= (5,1).
New answer posted
10 months agoContributor-Level 10
Reflected point of (2,1) about y-axis is (-2,1).
Reflected ray passes through (-2,1) and (5,3).
Equation: (y-1)/ (x+2) = (3-1)/ (5+2) = 2/7 ⇒ 2x - 7y + 11 = 0.
This is one directrix. Let the other be 2x - 7y + α = 0.
Distance between directrices = 2a/e = |11-α|/√53.
Distance from focus to directrix = a/e - ae = 8/√53.
a/e (1-e²) = 8/√53.
e=1/3. a/e (8/9) = 8/√53 ⇒ a/e = 9/√53.
2a/e = 18/√53 = |11-α|/√53.
|11-α| = 18.
11-α = 18 ⇒ α = -7.
11-α = -18 ⇒ α = 29.
Other directrix: 2x-7y-7=0 or 2x-7y+29=0.
New answer posted
10 months agoContributor-Level 10
S? : x² + y² - x - y - 1/2 = 0, C? : (1/2, 1/2), r? = √ (1/4)+ (1/4)+ (1/2) = 1.
S? : x² + y² - 4y + 7/4 = 0, C? : (0, 2), r? = √ (4 - 7/4) = 3/2.
S? : (x-2)² + (y-1)² ≤ r², C? : (2, 1).
A ∪ B ⊂ C means both circles S? and S? must be inside S?
Distance C? = √ (2-1/2)² + (1-1/2)²) = √ (9/4 + 1/4) = √10/2.
Condition: r ≥ C? + r? ⇒ r ≥ √10/2 + 1.
Distance C? = √ (2-0)² + (1-2)²) = √5.
Condition: r ≥ C? + r? ⇒ r ≥ √5 + 3/2.
√10/2 + 1 ≈ 1.58 + 1 = 2.58.
√5 + 3/2 ≈ 2.23 + 1.5 = 3.73.
So minimum r = (√10+2)/2 and (2√5+3)/2. We need the maximum of these two.
Let's recheck the question logic f
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