Math, asked by sumavaishnavi3528, 1 year ago

Find the equation of the ellipse which passes through the point, [4,1] and having its foci at [+-3,0].

Answers

Answered by yashvardhan5
1
sorry today could not
Answered by Sharad001
41

Question :-

→ Given above ↑

Answer :-

\boxed{ \to \sf \:   \frac{ {x}^{2} }{18}  +  \frac{ {y}^{2} }{9}  = 1} \:

To find :-

→ Equation of ellipse .

Step - by - step explanation :-

According to the question,

→ Ellipse is passes through (4,1)

Foci of required ellipse is at ( +3,0) and ( -3,0)

hence , c = 3

 \red{ \bf let \: required \:} \bf ellipse \: is \:  \\  \\  \to \sf  \frac{ {x}^{2} }{ {a}^{2} }  +  \frac{ {y}^{2} }{ {b}^{2} }  = 1 \\  \\  \sf \: we \: know \: that \:  \\  \:  \\  \to \sf \:  {c}^{2}  =  {a}^{2}   -   {b}^{2}  \\  \\  \to \sf {3}^{2}  =  {a}^{2}   -   {b}^{2}  \\  \\  \to \sf \: {a}^{2}   -   {b}^{2}  = 9 \: ........eq.(1) \\ \sf \:  or  \:   {a}^{2} = 9  +  {b}^{2}   \\  \\  \because \sf ellipse \: is \: passing \: through \: (4,1) \:    \\   \therefore \\  \to \sf \: \frac{ {(4)}^{2} }{ {a}^{2} }  +  \frac{ {(1)}^{2} }{ {b}^{2} }  = 1 \\  \\  \to \sf \frac{16}{ {b}^{2} + 9 }  +  \frac{1}{ {b}^{2} }  = 1 \\  \\  \to \sf 16 {b}^{2}  +  {b}^{2}  + 9 =  {b}^{4}  + 9 {b}^{2}  \\  \\  \to \sf {b}^{4}  + 8 {b}^{2}  - 9 = 0 \\  \\  \to \sf \:  {b}^{4}   +  9 {b}^{2}   -  {b}^{2}  - 9 = 0 \\  \\  \to \sf ( {b}^{2}  - 9)( {b}^{2}  + 1) = 0 \\  \\   \star \sf \: case \: (1) \\  \to \sf {b}^{2}  = 9 \\  \\  \to \boxed{ \sf \: b = 9} \\  \\  \star \sf \:  case \: (2) \\  \\  \to \sf {b}^{2}  =  - 1 \\  \because \sf real \:  number \: can \: not \: negative \\  \\  \therefore \:  \\  \to \sf {a}^{2}  =  {3}^{2}  + 9 \\  \\  \to  \boxed{\sf {a}^{2}  = 18}

hence required ellipse is

 \boxed{ \to \sf \:   \frac{ {x}^{2} }{18}  +  \frac{ {y}^{2} }{9}  = 1}

Hope it helps you .

Similar questions