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If the tangent to the ellipse ${x^2} + 2y = 1$ at point $P\left( {\frac{1}{{\sqrt 2 }},\frac{1}{2}} \right)$ meets the auxiliary circle at the points R and Q, then tangents to the circle at Q and R intersect at
$
  A)\left( {\frac{1}{{\sqrt 2 }},1} \right) \\
  B)\left( {1,\frac{1}{{\sqrt 2 }}} \right) \\
  C)\left( {\frac{1}{2},\frac{1}{2}} \right) \\
  D)\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right) \\
$

Answer
VerifiedVerified
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Hint: Here we will proceed the solution by finding tangent to the ellipse and chord of contact to the circle.
Given ellipse is${x^2} + 2y = 1$
It tangent is at point $P\left( {\frac{1}{{\sqrt 2 }},\frac{1}{2}} \right)$
We know that equation of tangent to the ellipse at point $p({x_1}{y_1})$is $x{x_1} + y{y_1} = 1$
If we consider the points in $P\left( {\frac{1}{{\sqrt 2 }},\frac{1}{2}} \right)$ as ${x_1}{y_1}$
Then equation of tangent to the ellipse at point P is
$
   \Rightarrow x\left( {\frac{1}{{\sqrt 2 }}} \right) + 2y\left( {\frac{1}{2}} \right) = 1 \\
    \\
$
$ \Rightarrow x + \sqrt 2 y = 1$ $ \to (1)$
Now QR is the chord of contact of circle${x^2} + {y^2} = 1$ at the point $T(h,k)$
Then, Chord of contact QR$ \equiv hx + ky = 1$$ \to (2)$
Here equation $(1)$ and$(2)$ represents two Straight lines
Now let us compare the coefficient in the ratio form, then we have
$ \Rightarrow $$\frac{h}{1} = \frac{k}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}$

From this we can say that Q and R intersect at point T (h, k)
Where (h, k) = $\left( {\frac{1}{{\sqrt 2 }},1} \right)$
$\therefore (h,k) = \left( {\frac{1}{{\sqrt 2 }},1} \right)$

Option A is Correct

NOTE: Here we will ignore finding the chord of contact to the circle as it is not directly mentioned in the question as tangent to the ellipse mentioned.