यदि अतिपरवलय 4 y ^{2}= x ^{2}+1 पर खींची गई स | vedclass

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Question

equation of hyperbola is :

$4{y^2} = {x^2} + 1 \Rightarrow  - {x^2} + 4{y^2} = 1$

$ \Rightarrow  - \frac{{{x^2}}}{{{1^2}}} + \frac{{{y

Vedclass · Accepted Answer

$x^2 - 4y^2 - 16 x^2 y^2 = 0$

Vedclass · Answer

equation of hyperbola is :

$4{y^2} = {x^2} + 1 \Rightarrow  - {x^2} + 4{y^2} = 1$

$ \Rightarrow  - \frac{{{x^2}}}{{{1^2}}} + \frac{{{y^2}}}{{{{\left( {\frac{1}{2}} ight)}^2}}} = 1$

$	herefore a = 1,b = \frac{1}{2}$

Now, tangent to the curve at point $\left( {{x_1},{y_1}} ight)$ is given by.

$4 	imes 2{y_1}\frac{{dy}}{{dx}} = 2{x_1}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2{x_1}}}{{8{y_1}}} = \frac{{{x_1}}}{{4{y_1}}}$

equation of tangent at $\left( {{x_1},{y_1}} ight)$ is 

$y = mx + c$

$ \Rightarrow y = \frac{{{x_1}{x_1}}}{{4{y_1}}}.x + c$

As tangent passes through ${{x_1},{y_1}}$

$	herefore {y_1} = \frac{{{x_1}{x_1}}}{{4{y_1}}} + c$

$ \Rightarrow C = \frac{{4y_1^2 - x_1^2}}{{4{y_1}}} = \frac{1}{{4{y_1}}}$

Therefore $y = \frac{{{x_1}}}{{4{y_1}}}x + \frac{1}{{4{y_1}}}$

$ \Rightarrow 4{y_1}y = {x_1}x + 1$  

which intersects at $A\left( {\frac{{ - 1}}{{{x_1}}},0} ight)$ and $y$ -axis at $B\left( {0,\frac{1}{{4{y_1}}}} ight)$

Let midpoint of $AB$ is $(h,k)$

$	herefore h = \frac{{ - 1}}{{2{x_1}}}$

$ \Rightarrow {x_1} = \frac{{ - 1}}{{2h}}\,\,\,\,\,\,{y_1} = \frac{1}{{8k}}$

Thus, $4{\left( {\frac{1}{{8k}}} ight)^2} = {\left( {\frac{{ - 1}}{{2h}}} ight)^2} + 1$

$ \Rightarrow \frac{1}{{16{k^2}}} = \frac{1}{{4{h^2}}} + 1$

$ \Rightarrow 1 = \frac{{16{k^2}}}{{4{h^2}}} + 16{k^2}\,\,\,\, \Rightarrow {h^2} = 4{k^2} + 16{h^2}k$

So, requatired equation is 

${x^2} - 4{y^2}16{x^2}{y^2} = 0$

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