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If the eccentricity of the standard hype...

If the eccentricity of the standard hyperbola passing through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is

A

`3x-2y=0`

B

`x-2y+8=0`

C

`2x-y-2=0`

D

`2x-3y+10=0`

Text Solution

Verified by Experts

Let the equation of standard hyperbola is
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 " (i)" `
Now, eccentricity of hyperbola is
` sqrt(1+(b^(2))/(a^(2)))=2 " " `(given)
`rArr a^(2)+b^(2)=4a^(2)`
`rArr b^(2)=3a^(2) " …(ii)" `
Since, hyperbola (i) passes through the point (4, 6)
` therefore (16)/(a^(2))-(36)/(b^(2))=1 " ...(iii)" `
On solving Eqs. (ii) and (iii), we get
`a^(2)=4 and b^(2) =12 " ...(iv)" `
Now, equation of tangent to hyperbola (i) at point (4, 6), is
`(4x)/(a^(2))-(6y)/(b^(2))=1`
`rArr (4x)/(4)-(6y)/(12)=1 " "`[from Eq.(iv)]
`rArr x-(y)/(2)=1 rArr 2x-y-2=0`
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