ઉપવલયની ઉકેન્દ્રિતા frac{1}{2} અને એક ના | vedclass

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Question

(a) There are two common tangents to the circle ${x^2} + {y^2} = 1$

and the hyperbola ${x^2} - {y^2} = 1.$

These are $x = 1$ and $x = - 1$

Ou

Vedclass · Accepted Answer

$\frac{{{{(x - 1/3)}^2}}}{{1/9}} + \frac{{{{(y - 1)}^2}}}{{1/12}} = 1$

Vedclass · Answer

(a) There are two common tangents to the circle ${x^2} + {y^2} = 1$

and the hyperbola ${x^2} - {y^2} = 1.$

These are $x = 1$ and $x = - 1$

Out of these, $x = 1$ is nearer to the point $P(1/2,1)$.

Thus a directrix of the required ellipse is $x = 1.$

If $Q(x,y)$ is any point on the ellipse,

then its distance from the focus is $QP = \sqrt {{{\left( {x - \frac{1}{2}} \right)}^2} + {{(y - 1)}^2}} $

and its distance from the directrix $x = 1$is $|x - 1|$.

By definition of ellipse, $QP = e|x - 1|$

$ \Rightarrow \sqrt {{{\left( {x - \frac{1}{2}} \right)}^2} + {{(y - 1)}^2}} = \frac{1}{2}|x - 1|$

==> $3{x^2} - 2x + 4{y^2} - 8y + 4 = 0$

or $\frac{{{{\left( {x - \frac{1}{3}} \right)}^2}}}{{1/9}} + \frac{{{{(y - 1)}^2}}}{{1/12}} = 1$.

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