Conic Sections
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New answer posted
4 weeks agoContributor-Level 9
Equation of chord of x² + y² = 25 with mid point (h, k) is xh + yk = h² + k².
Or, y = (-h/k)x + (h² + k²)/k.
If this touches the ellipse x²/9 + y²/16 = 1, then the condition for tangency c² = a²m² + b² must be satisfied.
Here, m = -h/k, c = (h²+k²)/k, a²=9, b²=16.
(h² + k²)/k)² = 9 (-h/k)² + 16
(h² + k²)²/k² = 9h²/k² + 16
⇒ (h² + k²)² = 9h² + 16k²
∴ Required locus (x² + y²)² = 9x² + 16y².
New answer posted
4 weeks agoContributor-Level 9
y² - 2x = 4√ (y² - 2x + 1)
Let y² - 2x = k.
k = 4√ (k+1) ⇒ k² = 16 (k+1) ⇒ k²-16k-16=0
(This appears to be from another problem. The provided solution text says:)
y² = 2x = 4 (1/2 x)
a > 2 * (1/2) ⇒ a > 1.
New answer posted
4 weeks agoContributor-Level 10
The tangent to the parabola y² = 4ax is y = mx + a/m.
For y² = 4x, a=1. So, the tangent is y = mx + 1/m.
The given line is y = mx + 4.
Comparing the two, 1/m = 4 ⇒ m = 1/4.
The line is y = (1/4)x + 4.
This line is also tangent to x² = 2by.
Substitute y into the parabola equation:
x² = 2b (1/4)x + 4)
x² = ( b/2 )x + 8b
x² - ( b/2 )x - 8b = 0.
For tangency, the discriminant (D) is zero.
D = (-b/2)² - 4 (1) (-8b) = 0.
b²/4 + 32b = 0.
b ( b/4 + 32) = 0.
b = 0 (not possible) or b/4 = -32 ⇒ b = -128.
New answer posted
4 weeks agoContributor-Level 10
Given 2ae = 6 and 2a/e = 12.
From these, ae = 3 and a/e = 6.
Multiplying the two equations: (ae) (a/e) = 3 * 6 => a² = 18.
We know that b² = a² (1 - e²) = a² - a²e² = 18 - (ae)² = 18 - 3² = 18 - 9 = 9.
The length of the latus rectum (L.R.) is 2b²/a.
L.R. = 2 * 9 / √18 = 18 / (3√2) = 6/√2 = 3√2.
New answer posted
4 weeks agoContributor-Level 10
The point of intersection of the ellipse x²/16 + y²/b² = 1 and the curve y² = 3x² lies on both.
Substitute y² = 3x² into the ellipse equation:
x²/16 + 3x²/b² = 1
x² (1/16 + 3/b²) = 1
x² (b² + 48) / 16b² = 1
x² = 16b² / (b² + 48).
For a solution to exist, we need x² > 0, which is true if b≠0.
The problem seems to have a condition missing or misinterpreted in the OCR. The provided solution also shows x² + y² = 4b, which might be another curve involved. Assuming the point lies on x²+y²=4b.
x² + 3x² = 4b => 4x² = 4b => x² = b.
Substitute x²=b into the ellipse equation: b/16 + 3b/b² = 1 (assuming y²=3b).
b/16 + 3/b = 1
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
The equation of the tangent to the ellipse x²/27 + y² = 1 at the point (3√3 cosθ, sinθ) is:
x (3√3 cosθ)/27 + y (sinθ)/1 = 1 ⇒ x/ (3√3) cosθ + y sinθ = 1.
To find the intercepts on the axes:
x-intercept (set y=0): x = 3√3 / cosθ = 3√3 secθ.
y-intercept (set x=0): y = 1 / sinθ = cosecθ.
The sum of the intercepts is z (θ) = 3√3 secθ + cosecθ.
To find the minimum value of z, we differentiate with respect to θ and set it to zero:
dz/dθ = 3√3 secθ tanθ - cosecθ cotθ = 0.
3√3 (sinθ/cos²θ) = cosθ/sin²θ.
3√3 sin³θ = cos³θ ⇒ tan³θ = 1/ (3√3).
⇒ tanθ = 1/√3.
Since θ ∈ (0, π/2), the solution
New answer posted
a month agoContributor-Level 10
The equation is for a hyperbola: x²/4 - y²/2 = 1.
The eccentricity e is given by e = √ (1 + b²/a²) = √ (1 + 2/4) = √6/2.
The focus F is at (ae, 0), which is (2 * √6/2, 0) = (√6, 0).
The equation of the tangent at a point P (x? , y? ) is xx? /a² - yy? /b² = 1.
The equation of the tangent at P is given as x - (√6 y)/2 = 1.
This tangent cuts the x-axis (y=0) at x=1, so Q is (1, 0).
The latus rectum is the line x = ae = √6.
To find the point R where the tangent intersects the latus rectum, we substitute x=√6 into the tangent equation:
√6 - (√6 y)/2 = 1 ⇒ √6 - 1 = (√6 y)/2 ⇒ y = 2 (√6 - 1)/√6.
The vertices o
New answer posted
a month agoContributor-Level 10
Given a triangle with inradius r and circumradius R.
The inradius r is calculated as r = |0 + 0 - 3| / √2 = 3 / √2.
From the geometry of the triangle, we have the relation r/R = sin (30°) = 1/2.
This gives R = 2r.
The question asks for the sum R + r, which is 2r + r = 3r.
Substituting the value of r, we get 3 * (3/√2) = 9/√2.
∴ R + r = 9/√2.
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