Question

10. If x/acos• + y/bsin • = 1 and x/ysin•- y/bcos•= 1, prove that x^2/a^2 + y^2/b^2= 2.​

Answer 1

Answer:

PROVED BELOW

Step-by-step explanation:

GIVEN:

$\frac{x}{acos\theta}+\frac{y}{bsin\theta}=1.....(i)\\ and\\\frac{x}{asin\theta}-\frac{y}{bcos\theta}=1........(ii)\\\\Now\\\\squaring\;and\;adding\;both\\we\\get\\=>[\frac{x}{acos\theta}+\frac{y}{bsin\theta}]^2+[\frac{x}{asin\theta}-\frac{y}{bcos\theta}]^2=[1]^2+[1]^2\\ \\=>\frac{x^2}{a^2cos^2\theta}+ \frac{y^2}{b^2sin^2\theta}+2[\frac{x}{acos\theta}][\frac{y}{bsin\theta}]+ \frac{x^2}{ a^2sin^2\theta}+ \frac{y^2}{b^2cos^2\theta}-2[\frac{x}{asin\theta}][\frac{y}{bcos\theta}] = 2$

$=>\frac{x^2}{a^2cos^2\theta}+ \frac{y^2}{b^2sin^2\theta}+\frac{x^2}{a^2sin^2\theta}+ \frac{y^2}{b^2cos^2\theta}+2[\frac{x}{acos\theta}][\frac{y}{bsin\theta}]-2[\frac{x}{asin\theta}][\frac{y}{bcos\theta}]=2$

$=>\frac{x^2}{a^2[cos^2\theta+sin^2\theta]}+ \frac{y^2}{b^2[sin^2\theta+cos^2\theta]}=2\\ \\\\=>\frac{x^2}{a^2}+ \frac{y^2}{b^2}=2$

Hence proved!

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