Question

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

Answer 1

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 .

Answer 2

Answer:u are a blogger on yt

Step-by-step explanation: