Conic Sections
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a month agoContributor-Level 10
Slope of tangent is 2, Tangent of hyperbola
x²/4-y²/2=1 at the point (x? , y? ) is xx? /4-yy? /2=1 (T=0)
Slope: x? /2y? =2 ⇒ x? =4y?
(x? , y? ) lies on hyperbola
⇒ x? ²/4-y? ²/2=1
From (1) and (2)
(4y? )²/4-y? ²/2=1 ⇒ 4y? ²-y? ²/2=1
⇒ 7y? ²=2 ⇒ y? ²=2/7
Now x? ²+5y? ² = (4y? )²+5y? ² = 21y? ² = 21*2/7=6
New answer posted
a month agoContributor-Level 10
|x|/2 + |y|/3 = 1
x²/4 + y²/9 = 1
Area of Ellipse = πab = 6π
Required area = π*2*3 - (Area of quadrilateral)
= 6π - 1/2*6*4
= 6π-12
= 6 (π-2)
New answer posted
a month agoContributor-Level 9
Hyperbola: 16 (x+1)² - 9 (y-2)² = 144. (x+1)²/9 - (y-2)²/16 = 1. Center (-1,2).
Foci (-1±ae, 2). a²=9, b²=16. e²=1+16/9=25/9, e=5/3. ae=5. Foci (4,2), (-6,2).
Centroid (h, k) of P, S, S': P (3secθ-1, 4tanθ+2).
h= (3secθ-1+4-6)/3 = secθ-1. k= (4tanθ+2+2+2)/3 = (4/3)tanθ+2.
(h+1)² - (3 (k-2)/4)² = 1. 16 (x+1)²-9 (y-2)²=16.
16x²+32x+16-9y²+36y-36=16. 16x²-9y²+32x+36y-36=0.
New answer posted
a month agoContributor-Level 9
Vertex (2,0), Focus (4,0). Parabola y²=4a (x-h) = 4 (2) (x-2) = 8 (x-2).
Tangents from O (0,0): T²=SS? (y (0)-4 (x+0)+16)²= (0-0+16) (y²-8x+16). No, this is for point on tangent.
Equation of tangent y=mx+a/m = m (x-2)+2/m. Passes through (0,0) so -2m+2/m=0 ⇒ m=±1.
Tangents y=x, y=-x. Points of contact S (4,4), R (4, -4).
Area of ΔSOR = ½ * base * height = ½ * 8 * 4 = 16.
New answer posted
a month 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
a month 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
a month 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)
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