The coordinates (2,3) and (1, 5) are the foci of an ellipse which passes through the origin. Then the equation of the

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tangent at the origin is $(3 \sqrt{2} - 5) x + (1 - 2 \sqrt{2}) y = 0$

tangent at the origin is $(3 \sqrt{2} + 5) x + (1 + 2 \sqrt{2}) y = 0$

normal at the origin is $(3 \sqrt{2} + 5) x - (1 + 2 \sqrt{2}) y = 0$

normal at the origin is $x (3 \sqrt{2} - 5) - y (1 - 2 \sqrt{2}) = 0$

answer is A, C.

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Detailed Solution

Tangent and normal are bisectors of $∠$ SPS''
Now, the equation of SP is $y = 3 x / 2$ and that of S'P is y= 5x.
Then their equations are
$\frac{3 x - 2 y}{\sqrt{13}} = \pm \frac{5 x - y}{\sqrt{26}} or 3 x - 2 y = \pm \frac{5 x - y}{\sqrt{2}}$
Therefore, the lines are
$(3 \sqrt{2} - 5) x + (1 - 2 \sqrt{2}) y = 0$
and $(3 \sqrt{2} + 5) x - (2 \sqrt{2} + 1) y = 0$
Now, (2,3)and (1, 5) lie on the same side of $(3 \sqrt{2} - 5) x +$
$(1 - 2 \sqrt{2}) y = 0$ which is the equation of tangent'