Question

Find the Centre,Foci, vertices and convertices of the hyperbole 16x²-224x+25y²+250y-191=0​

Answer 1

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

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

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

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

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

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

• According to given question :

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

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

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

Answer 2

$\huge\underline{\bf \orange{Aηsωeя :}}$

⟹ 16x²−224x+25y² +250y−191=0

⟹ (4x)² +28² −224x−28² +(5y)²+25² +250y−25² −191=0

⟹ (4x−28)² +(5y+25)² =191+ 625+ 784

⟹ (4x−28)² +(5y+25)² =1600

⟹ 1600(4x−28)² + 1600(5y+25)² =1

⟹ 160016(x−7)² + 160025(y+5)² =1

⟹ 100(x−7)² + 64(y+5)² =1

⟹ 10²(x−7)² + 8²(y+5)² =1

It is in the form of,

⟹ X²/a² + Y²/b² =1

As we know that,

⟹ x =0

⟹ x−7 =0

⟹ x= 7

⟹ Y =0

⟹ y+5= 0

⟹ y= −5

∴Center(7,−5)

As we know that,

⟹ X =0

⟹ x−7 =0

⟹ x =7

⟹ Y= ±b

⟹ y+ 5= ±8

⟹ y= −13 and 3

∴Vertex(7,−13) and(7,3)

As we know that,

⟹ a² =b² (e² −1)

⟹ 100=64(e² −1)

⟹ 64/100 =e² −1

⟹ 25/16 +1=e²

⟹ e= 41/4

As we know that,

⟹ x= 7

⟹ Y= ±be

⟹ y+ 5 = ±8× 41/4

⟹ y+ 5= ±2⁴¹

⟹ y= ±2⁴¹ −5

∴Foci(7,±2⁴¹ −5) Ans.