Question

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

• A. Tangent at origin is $(3\sqrt{2}-5)x+(1-2\sqrt{2})y=0$
• B. Tangent at origin is $(3\sqrt{2}+5)x+(1+2\sqrt{2})y=0$
• C. Normal at the origin is $(3\sqrt{2}+5)x-(1+2\sqrt{2})y=0$
• D. Normal at the origin is $x(3\sqrt{2}-5)-y(1-2\sqrt{2})=0$

Answer 1

Answer: (A), (C)

The correct options are
A Tangent at origin is $(3\sqrt{2}-5)x+(1-2\sqrt{2})y=0$
C Normal at the origin is $(3\sqrt{2}+5)x-(1+2\sqrt{2})y=0$

Let $P(0,0)$ be the origin

focii of the ellipse are $S(2,3)$ and $S^{\prime }(1,5)$

From reflection property, we know that Tangent and Normal are the angular bisectors of $\angle SP{S}^{\prime }$

Eqaution of $SP:y=\frac{3}{2}x$
Equation of $S^{\prime }P:y=\frac{5}{1}x$

Angle bisectors equations are::
$\frac{3x-2y}{\sqrt{13}}=\pm \frac{5x-y}{\sqrt{26}}$
We get two different equations with respect to signs

$(3\sqrt{2}-5)x+(1-2\sqrt{2})y=0\cdots \left(1\right)$

$(3\sqrt{2}+5)x-(2\sqrt{2}+1)y=0\cdots \left(2\right)$

We know thta normal will have both the focii on opposite sides

Substituting $S(2,3)$ and $S^{\prime }(1,5)$ in the equation $\left(1\right)$

We get the same sign, So equation $\left(1\right)$ is tangent equation

and equation $\left(2\right)$ is a normal equatiion