Find the equation to the cone whose vertex is the point (a,b,c) and whose generating lines
intersects the conic px2
+ qy2
= 1, z = 0.
To find the equation of the cone, we need to determine the equation of the generating lines first.
The generating lines of the cone intersect the conic section given by the equation px^2 + qy^2 = 1, z = 0. These lines lie on the surface of the cone.
Let's consider a generic point on the conic section with coordinates (x, y, z), where z is 0.
Substituting z = 0 in the equation of the conic, we get:
px^2 + qy^2 = 1
Now, let's consider the point (a, b, c) as the vertex of the cone.
The equation of the generating lines passing through (a, b, c) can be expressed as a parametric equation:
x = a + t(u)
y = b + t(v)
z = c + t
Here, (u, v) are the direction ratios of the generating lines, and t is a parameter.
We need to find the values of (u, v) that satisfy both the equation of the conic section and the parametric equation of the generating lines.
Substituting the expressions for x, y, and z in the equation of the conic, we get:
p(a + t(u))^2 + q(b + t(v))^2 = 1
Expand and simplify the equation:
pa^2 + 2pat(u) + pt^2(u)^2 + qb^2 + 2qbt(v) + qt^2(v)^2 = 1
Collecting terms with the same powers of t, we get:
(pt^2(u)^2 + qt^2(v)^2) + (2pa + 2qb)t(u) + (pa^2 + qb^2 - 1) = 0
Since this equation must hold for all values of t, each term must be equal to zero:
pt^2(u)^2 + qt^2(v)^2 = 0 (1)
2pa + 2qb = 0 (2)
pa^2 + qb^2 - 1 = 0 (3)
From equation (2), we can solve for u in terms of v:
u = -(pa + qb) / (qt(v))
Substituting this value of u in equation (1), we get:
p(-(pa + qb))^2 / (q^2t^2(v)^2) + qt^2(v)^2 = 0
Expand and simplify the equation:
p(a^2p^2 + 2apqb + q^2b^2) + q^2t^2(v)^4 = 0
Rearranging the terms, we get:
p(a^2p^2 + 2apqb + q^2b^2) = -q^2t^2(v)^4
Divide both sides by q^2p(a^2p + 2abq + q^2b^2):
(a^2p + 2abq + q^2b^2) / (p(a^2p + 2abq + q^2b^2)) = -t^2(v)^4
Simplify the expression:
1 / p = -t^2(v)^4
Now, we have a relationship between t and v. Since t is a parameter, we can consider v^4 as another parameter, say k.
Using this value of k, we can express v as:
v = ±(k)^(1/4)
Now, substitute the values of v and u back into the equation for x, y, and z:
x = a - t(pa + qb) / (q(k)^(1/4))
y = b ± (k)^(1/4)
z = c + t
Combining these equations, we get the final parametric equation for the generating lines of the cone:
x = a - t(pa + qb) / (q(k)^(1/4))
y = b ± (k)^(1/4)
z = c + t
The equation of the cone with vertex (a, b, c) and generating lines intersecting the conic px^2 + qy^2 = 1, z = 0 is given by this parametric equation.
Good
find the equation to the cone whose vertex is the point (a,b,c) and whose generating lines intersects the conic px2+by2+cz2=0
I think this will be a good place to start. The answer is given, and you can surely find your way there.
https://books.google.com/books?id=uwDfHbyl1IMC&pg=SL22-PA420&lpg=SL22-PA420&dq=Find+the+equation+to+the+cone+whose+vertex+is+the+point+(a,b,c)+and+whose+generating+lines+intersects+the+conic+px2+%2B+qy2&source=bl&ots=mE5TWchazF&sig=oOLM6wUqC9yTZcDi7LK36Dw63AI&hl=en&sa=X&ved=0ahUKEwi-wY3y8ebYAhVrjK0KHXUvDMQQ6AEIKTAA#v=onepage&q=Find%20the%20equation%20to%20the%20cone%20whose%20vertex%20is%20the%20point%20(a%2Cb%2Cc)%20and%20whose%20generating%20lines%20intersects%20the%20conic%20px2%20%2B%20qy2&f=false