If x-a -cos- y-b -sin- and z-c tan - then x 2 a 2- y 2 b 2-a z2c2b 1- z 2 c 2c z 2 c 2-1d 1-z 2 c 2

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Value of Standard Angle(30,60,90)

If x=a θcosϕ,...

Question

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ =

(a) $\frac{z^{2}}{c^{2}}$
(b) $1 - \frac{z^{2}}{c^{2}}$
(c) $\frac{z^{2}}{c^{2}} - 1$
(d) $1 + \frac{z^{2}}{c^{2}}$

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Solution

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Similar questions

(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = ?

(1 + \frac{z^{2}}{c^{2}})

(1 - \frac{z^{2}}{c^{2}})

(\frac{z^{2}}{c^{2}} - 1)

\frac{z^{2}}{c^{2}}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1 has

x = a sec θ cos ϕ, y = b sec θ sin ϕ

z = c tan θ

(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = (1 + \frac{z^{2}}{c^{2}})

(a e, 0)

x = \frac{a}{e}

b^{2} = a^{2} (1 - e^{2})

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