Conic Sections

Let an ellipse E: x²/a² + y²/b² = 1, a² > b², passes through (√3/2, 1) and has eccentricity 1/√3. If a circle, centered at focus F(α, 0), α > 0, of E and radius 2/√3 intersects E at two points P and Q, then PQ² is equal to:

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Contributor-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.

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5 months ago

Let the normal at a point $P$ on the curve $^{}^{}$ $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the $y - a x i s$ at $(0, \frac{3}{2})$ $(\frac{}{})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $| m |$ is equal to(2D)

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Contributor-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.

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5 months ago

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ $^{}^{}$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co - ordinate axes at $(\frac{}{})$ $(- \frac{1}{3 \sqrt[]{2}}, 0)$ and $(0, β)$ then $β$ is equal to:

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Contributor-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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5 months ago

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ $^{}$ , internally in the ratio $1 : 2$ , is

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Contributor-Level 9

Let point P be (2t, t²) and Q be (h,k).
h/3 = (2t)/3, k/3 = (-2+t²)/3

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5 months ago

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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5 months ago

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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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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5 months 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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alok kumar singh

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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6 months 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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6 months 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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New answer posted

6 months 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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