Question

Plzz anyone solve it plz​

Answer 1

☯ Given :-

$\bullet\sf \ x= ar\ sec \alpha\ cos\beta\\ \\ \bullet\sf y= br \ sec\alpha \ sin\beta\\ \\ \bullet\sf \ z= cr \ tan\alpha$

☯To Prove :-

$\it \dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=r^2$

☯ Solution:-

$\bullet\sf x= ar\ sec\alpha\ cos\beta \\ \\ \dashrightarrow\sf \bigg(\dfrac{x}{a}\bigg) =r\ sec\alpha \ cos\beta\\ \\ \\ \bullet\sf y= br\ sec\alpha\ sin\beta\\ \\\dashrightarrow\sf \bigg(\dfrac{y}{b}\bigg)= r \ sec\alpha\ sin\beta\\ \\ \\ \bullet\sf z= cr\ tan\alpha\\ \\ \dashrightarrow\sf \bigg(\dfrac{z}{c}\bigg)= r\ tan\alpha$

Now ,

$\dashrightarrow\it \bigg[\dfrac{x}{a}\bigg]^2+ \bigg[\dfrac{y}{b}\bigg]^2-\bigg[\dfrac{z}{c}\bigg]^2= r^2$

☯Taking LHS ,

• Put the values from the above ,

$\dashrightarrow\sf ( r\ sec\alpha \cos\beta)^2+(r\ sec\alpha\ sin\beta)^2- (r\ tan\alpha)^2\\ \\ \\\dashrightarrow\sf (r^2\ sec^2\alpha\ cos^2\beta)+(r^2\ sec^2\alpha\ sin^2\beta)-(r^2\ tan^2\alpha)\\ \\ \\\dashrightarrow\sf r^2\Big[( sec^2\alpha \cos^2\beta+sec^2\alpha\ sin^2\beta-tan^2\alpha)\Big]\\ \\ \\ \dashrightarrow\sf r^2\Big[sec^2\alpha(cos^2\beta+sin^2\beta)-tan^2\alpha \Big]\ \ \ \ \ \Big[\therefore\ sin^2\theta+cos^2\theta=1\Big]\\ \\ \\ \dashrightarrow\sf r^2 \Big[sec^2\alpha(1)-tan^2\alpha\Big]\\ \\ \\ \dashrightarrow\sf r^2(1)\ \ \ \ \ \ \Big[\therefore\ 1+tan^2\theta= sec^2\theta\Big]$

$\sf L.H.S\ \longrightarrow r^2\\ \\ \sf\ R.H.S\ \longrightarrow r^2$

$\underline{\boxed{\dashrightarrow{\red{\sf r^2=r^2}}}}$

$\sf\ \ \ \ \ \ L.H.S=R.H.S\ \ \ \ (hence\ proved!!)$