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प्रश्न
if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2 = 2`
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उत्तर १
`[x/a cos theta + y/b sin theta]^2 + [x/a sin theta - y/b cos theta] = (1)^2 + (1)^2`
`x^2/a^2 cos^2 theta + y^2/b^2 sin^2 theta (2xy)/(ab) cos theta sin theta = x^2/a^2 sin^2 theta + y^2/b^2 cos^2 theta - (2xy)/(ab) sin theta cos theta = 1 + 1`
`x^2/a^2 cos^2 theta + y^2/b^2 cos^2 theta + y^2/b^2 sin^2 theta = 2`
`cos^ theta [x^2/a^2 + y^2/b^2] + sin^2 theta(x^2/a^2 + y^2/a^2) = 2`
`x^2/a^2 + y^2/b^2` = (∴ `cos^2 theta + sin^2 theta = 1`)
उत्तर २
It is given that:
`x/a cos θ + y/b sin θ = 1` ....(A)
and `x/a sin θ - y/b cos θ = 1` ....(B)
On squaring equation (A), we get
`(x/a cos θ + y/b sin θ)^2 = (1)^2`
⇒ `x^2/a^2 cos^2 θ + y^2/b^2 sin^2 θ + 2 x/a . y/b sin θ. cos θ = 1` ....(c)
On squaring equation (B), we get
= `(x/a sin θ - y/b cos θ )^2 = (1)^2`
⇒ `x^2/a^2 sin^2 θ + y^2/b^2 cos^2 θ + 2 x/a . y/b sin θ. cos θ = 1` ....(D)
Adding (C) and (D), we get,
⇒ `x^2/a^2 cos^2 θ + y^2/b^2 sin^2 θ + 2 x/a . y/b sin θ. cos θ + x^2/a^2 sin^2 θ + y^2/b^2 cos^2 θ + 2 x/a . y/b sin θ. cos θ = 1 + 1`
⇒ `x^2/a^2 sin^2 θ + y^2/b^2cos^2 θ-(4xy)/"ab" sin^2 θ + cos^2 θ = 2`
⇒ `x^2/a^2 xx 1 + y^2/b^2 xx 1 = 2`
⇒ `x^2/a^2 + y^2/b^2 = 2`
Hence proved.
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