Question

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

Class 11th Conic Sections .
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Answer 1

Let the equation of standard hyperbola is

$\dfrac{ {x}^{2} }{a {}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

Now ,

Given that is ecentricity is 2.

So,

$\sqrt{ 1 + \dfrac{ {b}^{2} }{ {a}^{2} } } = 2 \\ \\ {a {}^{2} + {b}^{2} } = 4 {a}^{2} \\ \\ \pink{\huge \boxed{ {b}^{2} = 3 {a}^{2} }}....(1)$

Also Hyperbola passes through the point

(4,6).

$\huge \pink {\boxed{{\dfrac{16}{ {a}^{2} } - \dfrac{36}{ {b}^{2} } = 1 }}}....(2)$

Solving (1) & (2) 》 ^_^

${a}^{2} = 4 \: \: and \: \: {b}^{2} = 12$

So equation of tangent to hyperbola at point (4,6)

is

$\dfrac{4x}{ {a}^{2} } - \dfrac{6y}{ {b}^{2} } = 1 \\ \\ \dfrac{4x}{4} - \frac{6y}{12} = 1 \\ \\ 2x - y = 2$

Answer 2

Correct Question:-)

If ecentricity of a standard

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

Solution:-)

Firstly Let us consider that equation of hyperbola is :-

$\dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

Which is the general equation of a standard hyperbola.

We know that formula of ecentricity of hyperbola is

$\bold{e = \sqrt{1 + \dfrac{ {a}^{2} }{ {b}^{2} } } }$

So According to the given question

$\sqrt{1 + \dfrac{ {a}^{2} }{ {b}^{2} } } = 2 \\ \\ \: squaring \: \: \: \: b.s. \\ \\ \implies1 + \dfrac{ {a}^{2} }{ {b}^{2} } = 4 \\ \\ \implies \red{\boxed{ \boxed{{ \bold{3 {a}^{2} = b}}}}} \: \: \: \: \: {eq}^{n} (i)$

Also jeevan hyperbola passes through ( 4 , 6 )

So,

it will satisfy the equation of hyperbola

i.e.

$\red{ \boxed{ \boxed { \bold{\dfrac{16}{ { a}^{2} } - \dfrac{36}{ {b}^{2} } = 1}}}} \: \: \: \: \: \: \: \: {eq}^{n} (ii)$

Solving Both equations (i) and (ii) Simultaneously

$\red{ \boxed{ \boxed { \bold{ {a = 4}^{2} }}}} \: \: \: and \: \: \:\red{ \boxed{ \boxed { \bold{ {b}^{2} = 12}}}}$

Now

equation of tangent will be

$\dfrac{4x}{ {a}^{2} } - \dfrac{6y}{ {b}^{2} } = 1 \\ \\ \frac{4x}{4} - \frac{6y}{12} = 1 \\ \\ x - \frac{y}{2} = 1 \\ \\\purple{ \boxed{ \boxed { \bold{2x - y - 2 = 0}}}}$