Question

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.​

Answer 1

Solution:

The guiding curve is

px² + qy² = 1, z = 0

Let, the straight line through (a, b, c) be

(x - a) / l = (y - b) / m = (z - c) / n ..... (1)

This line meets the plane z = 0

Then, (x - a) / l = (y - b) / m = - c / n ... (1)

This gives x = a - cl/n , y = b - cm/n

The point (a - cl/n, b - cm/n, 0) lies on the curve px² + qy² = 1, then

p (a - cl/n)² + q (b - cm/n)² = 1

or, p (an - cl)² + q (bn - cm)² = n²

or, p {a (z - c) - c (x - a)}² + q {b (z - c) - c (y - b)}² = (z - c)²

or, p (az - cx)² + q (bz - cy)² = (z - c)² ,

which is the equation of the required cone.

Answer 2

Given: px2 + qy2 = 1, z=0

To find: Equation of cone

Solution:

• We have given the conic equation as px2 + qy2 = 1 and z = 0, lets consider a straight line to be in terms of i, j and k.

• So the straight line through ( i,j,k ) will be:

               (x - i) / l = (y - j) / m = (z - k) / n     ........(i)

• Now we have given that z = 0, so put it in the above equation, we get:

              (x - i) / l = (y - j) / m = - k / n ... (1)

• On further solving we get:

              x = a - il/n and y = b - mk/n

• Now the point (a - kl/n, b - km/n, 0) lies on the curve px² + qy² = 1, then

              p (i - kl/n)² + q (j - km/n)² = 1

              p (in - kl)² + q (jn - km)² = n²

              p {i (z - k) - k (x - i)}² + q {j (z - k) - k (y - j)}² = (z - k)²

             p (iz - kx)² + q (jz - ky)² = (z - k)²

Answer:

• So the required equation is   p (iz - kx)² + q (jz - ky)² = (z - k)² .