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If the tangents drawn to the hyperbola $4y^{2}=x^{2}+1$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

Option 1)
$x^{2}-4y^{2}+16x^{2}y^{2}=0$

Option 2)
$x^{2}-4y^{2}-16x^{2}y^{2}=0$

Option 3)
$4x^{2}-y^{2}+16x^{2}y^{2}=0$

Option 4)
$4x^{2}-y^{2}-16x^{2}y^{2}=0$

Answers (1)

best_answer

As we learned

Mid-point formula -

$x= \frac{x_{1}+x_{2}}{2}$

$y= \frac{y_{1}+y_{2}}{2}$

- wherein

If the point P(x,y) is the mid point of line joining A(x₁,y₁) and B(x₂,y₂) .

Locus -

Path followed by a point p(x,y) under given condition (s).

- wherein

It is satisfied by all the points (x,y) on the locus.

Equation of Tangent to Hyperbola -

$\frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}= 1$

- wherein

For the Hyperbola

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

$P\left ( x_{1} ,y_{1}\right )$

Hyperbola is $\frac{y^{2}}{\frac{1}{4}}-\frac{x^{2}}{1}=1$

$4\beta y=\alpha x+1$

$\alpha x-4\beta y+1=0$

$\alpha =\frac{-1}{2h},\: \beta =\frac{1}{8k}$

Also , $4\beta ^{2}=\alpha ^{2}+1$

$\frac{4}{64k^{2}}=\frac{1}{4h^{2}}+1$

$\frac{1}{16k^{2}}=\frac{1}{4h^{2}}+1$

$h^{2}-4k^{2}-16h^{2}k^{2}=0$

$\Rightarrow x^{2}-4y^{2}-16x^{2}y^{2}=0$

Option 1)

$x^{2}-4y^{2}+16x^{2}y^{2}=0$

Option 2)

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

Option 3)

$4x^{2}-y^{2}+16x^{2}y^{2}=0$

Option 4)

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

Posted by

Himanshu

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