Conic Sections

Let P be a point on the parabola, y² = 12x and N be the foot of the perpendicular drawn from P on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is 4/3, then:

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Contributor-Level 10

Let P = (3t², 6t); N = (3t²,0)
M = (3t², 3t)
Equation of MQ: y = 3t
∴ Q = (3/4 t², 3t)
Equation of NQ
y = ( 3t / (3/4 t² - 3t²) ) (x - 3t²)
y - intercept of NQ = 4t = 4/3 ⇒ t = 1/3
∴ MQ = 9/4 t² = 1/4
PN = 6t = 2

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5 months ago

A hyperbola having the transverse axis of length √2 has the same foci as that of the ellipse 3x² + 4y² = 12, then this hyperbola does not pass through which of the following points?

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Contributor-Level 10

Ellipse: x²/4 + y²/3 = 1
eccentricity = √ (1 - 3/4) = 1/2
∴ foci = (±1,0)
For hyperbola, given 2a = √2 ⇒ a = 1/√2
∴ hyperbola will be x²/ (1/2) - y²/b² = 1
eccentricity = √ (1 + 2b²)
∴ foci = (±√ ( (1+2b²)/2 ), 0)
∴ Ellipse and hyperbola have same foci
√ ( (1+2b²)/2 ) = 1
⇒ b² = 1/2
∴ Equation of hyperbola: x²/ (1/2) - y²/ (1/2) = 1
⇒ x² - y² = 1/2
Clearly, (√3/2, 1/2) does not lie on it.

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The area (in sq. units) of an equilateral triangle inscribed in the parabola y²=8x, with one of its vertices on the vertex of this parabola, is

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Contributor-Level 9

Kindly consider the following Image

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5 months ago

For some θ ∈ (0, π/2), if the eccentricity of the hyperbola x² - y²sec²θ = 10 is √5 times the eccentricity of the ellipse x²sec²θ + y² = 5, then the length of the latus rectum of the ellipse is

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Contributor-Level 9

Kindly consider the following Image

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A line parallel to the straight line 2x-y=0 is tangent to the hyperbola x²/4 - y²/2 = 1 at the point (x₁,y₁). Then x₁²+5y₁² is equal to

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Contributor-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

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5 months ago

Area (in sq. units) of the region outside |x|/2 + |y|/3 = 1 and inside the ellipse x²/4 + y²/9 = 1 is

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Contributor-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)

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5 months ago

The locus of the centroid of the triangle formed by any point P on this hyperbola 16x² - 9y² + 32x + 36y - 164 = 0, and its foci is:

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Contributor-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.

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Let a parabola P be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from O(0, 0) to the parabola P which meet P at S and R, then the area (in sq. units) of ΔSOR is equal to:

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