After receiving two successive raises, hursh's salary became equal to 15/8 times of his initial salary. by how much percent was the salary raised the first time if the second raise was twice as high (in percent) as the first?
Answers
Explanation:
∴Center=(7,−5)
\green{\tt{\therefore{Vertices=(7,-13)\:and\:(7,3)}}}∴Vertices=(7,−13)and(7,3)
\green{\tt{\therefore{Foci=(7,\pm2\sqrt{41}-5)}}}∴Foci=(7,±2
41
−5)
\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}
Step−by−stepexplanation:
\begin{gathered}\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 =?\end{gathered}
Given:
:⟹16x
2
−224x+25y
2
+250y−191=0
ToFind:
:⟹Center=?
:⟹Foci=?
:⟹Vertices=?
:⟹Co−vertices=?
• According to given question :
\begin{gathered}\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\end{gathered}
Asweknowthat
:⟹16x
2
−224x+25y
2
+250y−191=0
:⟹(4x)
2
+28
2
−224x−28
2
+(5y)
2
+25
2
+250y−25
2
−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
:⟹
1600
16(x−7)
2
+
1600
25(y+5)
2
=1
:⟹
100
(x−7)
2
+
64
(y+5)
2
=1
:⟹
10
2
(x−7)
2
+
8
2
(y+5)
2
=1
\begin{gathered}\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\end{gathered}
It is in the form of
:⟹
a
2
X
2
+
b
2
Y
2
=1
Asweknowthat
:⟹x=0
:⟹x−7=0
:⟹x=7
:⟹Y=0
:⟹y+5=0
:⟹y=−5
∴Center(7,−5)
Asweknowthat
:⟹X=0
:⟹x−7=0
:⟹x=7
\begin{gathered}\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)}\end{gathered}
:⟹Y=±b
:⟹y+5=±8
:⟹y=−13and3
∴Vertex(7,−13)and(7,3)
Asweknowthat
:⟹a
2
=b
2
(e
2
−1)
:⟹100=64(e
2
−1)
:⟹
64
100
=e
2
−1
:⟹
16
25
+1=e
2
:⟹e=
4
41
Asweknowthat
:⟹x=7
:⟹Y=±be
:⟹y+5=±8×
4
41
:⟹y+5=±2
41
:⟹y=±2
41
−5
∴Foci(7,±2
41
−5)