Question

Find the envelope of the plane. Lx+my+nz = p when a²l²+b²m²+c²n² = p²

Answer 1

Explanation:

find the envelope of the plane LX +my+nz=p when a

Answer 2

Answer:

The envelope of the plane is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 2z$.

Explanation:

Let l/p = λ, m/p = µ, and n/p = ν

The equation of plane becomes

f ≡ λx + μy + νz - 1 = 0 ...(i), when

a^2λ^2 + b^2μ^2 + 2ν = 0 ...(ii)

Differentiating equations (i) and (ii), we get

xdλ + ydμ + zdν = 0

and a^2λdλ + b^2μ^2dμ + dν = 0

Comparing the above two equations, we get

(a^2 λ)/x = (b^2 μ)/y = 1/z = (a^2λ^2 + b^2μ^2 + ν)/(λx + μy + νz) = -ν/1

Therefore,

λ = $\frac{x}{a^2z}$, μ = $\frac{y}{b^2z}$, and ν = $-\frac{1}{z}$

Substituting λ, μ, and ν in (i), we get

(x^2/a^2) + (y^2/b^2) = 2z

Therefore, the envelope of the plane lx + my + nz = p when a²l²+b²m²+c²n² = p² is  $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 2z$.

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