Question

help please.............​

Answer 1

Step-by-step explanation:

here is the answer hope you guys understand and sorry for my bad handwriting

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

$:\mapsto\sf{\red{\:\dfrac{x}{a\cos \theta}\:=\:\dfrac{y}{b\sin \theta}----(1)}} \\ \\ \\ :\mapsto\sf{\red{\:\dfrac{ax}{\cos \theta}\:-\:\dfrac{by}{\sin \theta}\:=\:a^2-b^2----(2)}}$

$\Large{\underline{\mathfrak{\bf{To\:prove}}}}$

$:\mapsto\sf{\pink{\:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\:=\:1}}$

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

By,equation(2)

$:\mapsto\sf{\:\dfrac{ax}{\cos \theta}\:-\:\dfrac{by}{\sin \theta}\:=\:a^2-b^2} \\ \\ \\ \small\sf{\green{\:we\:can\:write\:in\:this\:type}} \\ \\ \\ :\mapsto\sf{\:a^2(\dfrac{x}{a\cos \theta})-\dfrac{by}{\sin \theta}\:=\:a^2-b^2} \\ \\ \\ \small\sf{\green{\:\:\:\:keep\:value\:by\:equ(1)}} \\ \\ \\ :\mapsto\sf{\:a^2(\dfrac{y}{b\sin \theta})-\dfrac{by}{\sin \theta}\:=\:a^2-b^2} \\ \\ \\ :\mapsto\sf{\:\dfrac{y}{b\sin \theta}\left(a^2-b^2\right)\:=\:(a^2-b^2)} \\ \\ \\ :\mapsto\sf{\:\dfrac{y}{b\sin \theta}\:=\:\cancel{\dfrac{(a^2-b^2)}{(a^2-b^2)}}} \\ \\ \\ :\mapsto\sf{\orange{\:y\:=\:b\sin \theta}}$

keep value of y in equ(1),

$:\mapsto\sf{\:\cancel{\dfrac{a\cos \theta}{a \cos \theta}}\:=\:\dfrac{y}{b \sin \theta}} \\ \\ \\ :\mapsto\sf{\orange{\:y\:=\:b\sin \theta}}$

We prove here,

$:\mapsto\sf{\pink{\:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\:=\:1}}$

Take, L.H.S.,

$:\mapsto\sf{\:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}} \\ \\ \\ \small\sf{\green{\:\:\:keep\:value\:of\:x\:and\:y}} \\ \\ \\ :\mapsto\sf{\:\dfrac{a^2\cos^2 \theta}{a^2}+\dfrac{b^2\sin^2 \theta}{b^2}} \\ \\ \\ :\mapsto\sf{\:\sin^2 \theta+\cos^2 \theta} \\ \\ \\ \small\sf{\green{\:\:\:\star\sin^2 \theta+\cos^2 \theta\:=\:1}} \\ \\ \\ :\mapsto\sf{\red{\:1}}$

= R.H.S.

That's proved.