Saved Bookmarks
| 1. |
If x = a (1 + cos θ cos ϕ), y = b (1 + cos θ sin ϕ) and z = c (1 + sin θ), then which of the following is correct?(a) \(\bigg(\frac{x-a}{a}\bigg)^2\) + \(\bigg(\frac{y-b}{b}\bigg)^2\) + \(\bigg(\frac{z-c}{c}\bigg)^2\)(b) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)(c) x2 + y2 + z2 = a2 + b2 + c2 (d) \(\frac{(x-a)^2}{a}+\frac{(y-b)^2}{b}+\frac{(z-c)^2}{c}=1\) |
|
Answer» (a) \(\bigg(\frac{x-a}{a}\bigg)^2\) + \(\bigg(\frac{y-b}{b}\bigg)^2\) + \(\bigg(\frac{z-c}{c}\bigg)^2\) Given, \(x\) = a (1 + cos θ cos ϕ) ⇒ \(\frac{x}{a}\) = 1+ cos θ cos ϕ ⇒ \(\frac{x}{a}\) - 1 cos θ cos ϕ ⇒ \(\frac{x-a}{a}\) = cos θ cos ϕ ...(i) Similarly, y = b (1 + cos θ sin ϕ ) ⇒ \(\frac{y-b}{b}\) = cos θ sin ϕ ...(ii) z = c (1+sin θ) ⇒ \(\frac{z-c}{c}\) = sin θ ...(iii) Squaring eqns (i), (ii) and (iii) and adding, we get \(\bigg(\frac{x-a}{a}\bigg)^2\) + \(\bigg(\frac{y-b}{b}\bigg)^2\) + \(\bigg(\frac{z-c}{c}\bigg)^2\) = cos2 θ cos2 ϕ + cos2 θ sin2 ϕ + sin2 θ = cos2 θ (cos2 ϕ + sin2 ϕ) + sin2 θ = cos2 θ + sin2 θ = 1. |
|