Question

Find the foci and vertices of the hyperbola 4x - 24x - 25y + 250y - 489 = 0​

Answer 1

$\textbf{Given:}$

$4x^2-24x-25y^2+250y-489= 0$

$4(x^2-6x)-25(y^2-10y)-489= 0$

$\text{To make them, intems of perfect square}$

$4((x^2-6x+9)-9)-25((y^2-10y+25)-25)-489= 0$

$4((x-3)^2-9)-25((y-5)^2-25)-489= 0$

$4(x-3)^2-36-25(y-5)^2+625-489= 0$

$4(x-3)^2-25(y-5)^2+100= 0$

$25(y-5)^2-4(x-3)^2=100$

$\text{Divide bothsides by 100, we get}$

$\displaystyle\frac{(y-5)^2}{4}-\frac{(x-3)^2}{25}=1$

$\text{This equation of the form \displaystyle\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1 }$

$\text{Here, a^2=4 and b^2=25 }$

$\text{comparing, we get Centre (h,k)=(3,5)}$

$\bf\;ae=\sqrt{a^2+b^2}$

$\implies\;ae=\sqrt{4+25}$

$\implies\;ae=\sqrt{29}$

$\text{Foci are (h,k-ae) and (h,k+ae)}$

$\implies(3,5-\sqrt{29})\;\text{and}\;(3,5+\sqrt{29})$

$\text{Vertices are (h,k-a) and (h,k+a)}$

$\implies(3,5-2)\;\text{and}\;(3,5+2})$

$\implies(3,3)\;\text{and}\;(3,7})$

$\therefore\textbf{Foci are \bf\,(3,5-\sqrt{29}) and \bf\,(3,5+\sqrt{29}) and}$

$\textbf{Vertices are (3,3) and (3,7)}$

Find more:

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$36x^2+144y^2-36x-96y-119=0$

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Center=(7,-5)}}}$

$\green{\tt{\therefore{Vertices=(7,-13)\:and\:(7,3)}}}$

$\green{\tt{\therefore{Foci=(7,\pm2\sqrt{41}-5)}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies 16 {x}^{2} - 224x + 25 {y}^{2} + 250y - 191 = 0 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Center = ?\\ \\ \tt: \implies Foci=? \\ \\ \tt: \implies Vertices =? \\ \\ \tt: \implies Co-vertices =?$

• According to given question :

$\bold{As \: we \: know \:that} \\ \tt: \implies 16 {x}^{2} - 224x + 25 {y}^{2} + 250y - 191 = 0 \\ \\ \tt: \implies {(4x)}^{2} + {28}^{2} - 224x - {28}^{2} + {(5y )}^{2} + {25}^{2} + 250y - {25}^{2} - 191 = 0 \\ \\ \tt: \implies {(4x - 28)}^{2} +( 5 {y + 25)}^{2} = 191 + 625 + 784 \\ \\ \tt: \implies {(4x - 28)}^{2} + {(5y + 25)}^{2} = 1600 \\ \\ \tt: \implies \frac{ {(4x - 28)}^{2} }{1600} + \frac{ {(5y + 25)}^{2} }{1600} = 1 \\ \\ \tt: \implies \frac{16(x - 7)^{2} }{1600} + \frac{25( {y + 5)}^{2} }{1600} = 1 \\ \\ \tt: \implies \frac{(x - 7 )^{2} }{100} + \frac{(y + 5)^{2} }{64} = 1 \\ \\ \tt: \implies \frac{ {(x - 7)}^{2} }{ {10}^{2} } + \frac{ {(y + 5)}^{2} }{ {8}^{2} } = 1$

$\text{It \: is \: in \: the \: form \: of} \\ \\ \tt: \implies \frac{ {X}^{2} }{ {a}^{2} } + \frac{ {Y}^{2} }{ {b}^{2} } = 1 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies x = 0 \\ \\ \tt: \implies x - 7 = 0 \\ \\ \tt: \implies x = 7 \\ \\ \tt: \implies Y = 0 \\ \\ \tt: \implies y + 5 = 0 \\ \\ \tt: \implies y = - 5 \\ \\ \green{\tt \therefore Center(7,- 5)} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies X = 0 \\ \\ \tt: \implies x - 7 = 0 \\ \\ \tt: \implies x = 7$

$\tt: \implies Y = \pm b \\ \\ \tt: \implies y + 5 = \pm 8 \\ \\ \tt: \implies y = -13 \: and \: 3 \\ \\ \green{\tt \therefore Vertex(7,-13) \: and \: (7,3)} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {a}^{2} = {b}^{2} ( {e}^{2} - 1) \\ \\ \tt: \implies 100 = 64( {e}^{2} - 1) \\ \\ \tt: \implies \frac{100}{64} = {e}^{2} - 1 \\ \\ \tt: \implies \frac{25}{16} + 1 = {e}^{2} \\ \\ \tt: \implies e = \frac{ \sqrt{41} }{4} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies x = 7 \\ \\ \tt: \implies Y = \pm be \\ \\ \tt: \implies y + 5 = \pm 8 \times \frac{ \sqrt{41} }{4} \\ \\ \tt: \implies y + 5= \pm 2 \sqrt{41} \\ \\ \tt: \implies y = \pm 2 \sqrt{41} - 5 \\ \\ \green{\tt \therefore Foci (7,\pm 2\sqrt{41} - 5)}$