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अगर xaybtanθ=1frac{x}{a}-frac{y}{b} an heta=1 और xatanθ+yb=1frac{x}{a} an heta+frac{y}{b}=1 , तो x2a2+y2b2frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}} का मूल्य है

a

2 cos2θ

b

sec2θ

c

2 sec2θ

d

cos2θ

Answer : Option A
Explanation :

xaybtanθ=1frac{x}{a}-frac{y}{b} an heta=1

xatanθ+yb=1frac{x}{a} an heta+frac{y}{b}=1

समीकरणों (i) और (ii) का वर्ग करने और जोड़ने पर,

x2a22xyabtanθ+y2b2tan2θ+x2a2tan2θfrac{x^{2}}{a^{2}}-frac{2 x y}{a b} an heta+frac{y^{2}}{b^{2}} an ^{2} heta+frac{x^{2}}{a^{2}} an ^{2} heta

+2xyabtanθ+y2b2=1+1+frac{2 x y}{a b} an heta+frac{y^{2}}{b^{2}}=1+1

x2a2+x2a2tan2θ+y2b2tan2θ+y2b2=2Rightarrow frac{x^{2}}{a^{2}}+frac{x^{2}}{a^{2}} an ^{2} heta+frac{y^{2}}{b^{2}} an ^{2} heta+frac{y^{2}}{b^{2}}=2

x2a2(1+tan2θ)+y2b2(tan2θ+1)=2Rightarrow frac{x^{2}}{a^{2}}left(1+ an ^{2} heta ight)+frac{y^{2}}{b^{2}}left( an ^{2} heta+1 ight)=2

x2a2sec2θ+y2b2sec2θ=2Rightarrow frac{x^{2}}{a^{2}} sec ^{2} heta+frac{y^{2}}{b^{2}} sec ^{2} heta=2

x2a2+y2b2=2sec2θ=2cos2θRightarrow frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=frac{2}{sec ^{2} heta}=2 cos ^{2} heta

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