Conic Sections

If the line y=mx+c is a common tangent to the hyperbola x²/100 - y²/64 = 1 and the circle x²+y²=36, then which one of the following is true?

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Contributor-Level 10

c² = 36(1+m²)
c² = 100m² - 64
100m² – 64 = 36 + 36m²

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For $a > 0$ , let the curves $C_{1} : y^{2} = a x$ $_{}^{}$ and $C_{2} : x^{2} =$ $_{}^{}$ ay intersect at origin $O$ and a point $P$ . Let the line $x = b (0 < b < c)$ intersect the chord $O P$ and the $x$ - axis at points $Q$ and $R$ , respectively. If the line $x = b$ bisects the area bounded by the curves, $C_{1}$ $_{}$ and $_{}$ , and the area of $\frac{}{}$ $? O Q R = \frac{1}{2}$ , then 'a' satisfies the equation:

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Raj Pandey

Contributor-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

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a month ago

If $3 x + 4 y = 12 \sqrt[]{2}$ is a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$ for some $a \in R$ , then the distance between the foci of the ellipse is:

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Contributor-Level 10

$y = - \frac{3 x}{4} + 3 \sqrt[]{2}$ line is tangent to ellipse

$∴ c^{2} = a^{2} m^{2} + b^{2}$

$18 = \frac{9 a^{2}}{16} + 9$

$a^{2} = 16$

$∴ e^{2} = 1 - \frac{b^{2}}{a^{2}}$

$e = \frac{\sqrt[]{7}}{4}$

Distance between focii $= 2 a e$

$= 2 \sqrt[]{7}$

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a month ago

Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to:

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Contributor-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).

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a month ago

A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of any ellipse with eccentricity 1/3 and the distance of the nearer focus from this directrix is 8/√53, then the equation of the other directrix can be:

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Contributor-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.

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a month ago

Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0.
If y(-2/3 logₑ2) = α logₑ2, then the value of α is equal to:

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Contributor-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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a month ago

If a tangent to the ellipse x² + 4y² = 4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:

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Contributor-Level 10

Equation of tangent of P (2cosθ, sinθ) is
(cosθ)x + (2sinθ)y = 4
Solving equation of tangent with equation of tangents at major axis ends, i.e. x = -2 and x = 2
For point 'B' (at x=-2):
-2cosθ + 2sinθ y = 4 ⇒ y = (2+cosθ)/sinθ
B (-2, (2+cosθ)/sinθ)
For point 'C' (at x=2):
2cosθ + 2sinθ y = 4 ⇒ y = (2-cosθ)/sinθ
C (2, (2-cosθ)/sinθ)
Now BC is the diameter of circle
Equation of circle: (x+2) (x-2) + (y - (2+cosθ)/sinθ) (y - (2-cosθ)/sinθ) = 0
x²-4 + y² - (4/sinθ)y + (4-cos²θ)/sin²θ = 0
Check if (√3, 0) satisfies this:
(√3)²-4 + 0 - 0 + (4-cos²θ)/sin²θ = -1 + (3+sin²θ)/sin²θ = -1 + 3/sin²θ + 1 = 3/sin²

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a month ago

Let A(1, 4) and B(1, -5) be two points. Let P be a point on the circle ${(x - 1)}^{2} + {(y - 1)}^{2} = 1$ such that ${(P A)}^{2} + {(P B)}^{2}$ have a maximum value, then the points P, A and B lie on:

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Contributor-Level 10

$P A^{2} + P B^{2} = c o s^{2} θ + {(s i n θ - 3)}^{2} + c o s^{2} θ + {(s i n θ + 6)}^{2}$

= 2 cos² θ + 2 sin² θ + 6 sin q + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin θ = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.

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a month ago

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 36,$ which is outside the parabola y² = 9x is:

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Contributor-Level 10

x² + y² = 36 ……(i)

y² = 9x .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A = \int_{0}^{3} (\sqrt[]{36 - x^{2}} - 3 \sqrt[]{x}) d x$

$= {[\frac{x \sqrt[]{36 - x^{2}}}{2} + 18 s i n^{- 1} \frac{x}{6}]}_{0}^{3} - 3 . {\frac{x^{3 / 2}}{3 / 2}]}_{0}^{3}$

$= 3 π - \frac{3 \sqrt[]{3}}{2}$

Required area = $= \frac{1}{2} π {(6)}^{2} + 2 (3 π - \frac{3 \sqrt[]{3}}{2}) = (24 π - 3 \sqrt[]{3}) s q . u n i t$

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a month ago

The locus of the mid- point of the line segment joining the focus of the parabola y² = 4ax to a moving point of the parabola, is another parabola whose directrix is:

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