Question

Consider equation of ellipses as:
$E_1:\frac{x^2}{25}+\frac{y^2}{9}=1$
$E_2:\frac{x^2}{9}+\frac{y^2}{4}=1$
$E_3:\frac{x^2}{9}+\frac{y^2}{25}=1$
Which of the following is/are true?

• A. Intersection of $E_1$ and $E_2$ have four concyclic points.
• B. Intersection of $E_1$ and $E_3$ have four concyclic points.
• C. Intersection of $E_2$ and $E_3$ have four concyclic points.
• D. None of these

Answer 1

The correct option is B Intersection of $E_1$ and $E_3$ have four concyclic points.

$E_1=\frac{x^2}{25}+\frac{y^2}{9}=1$

$E_2=\frac{x^2}{9}+\frac{y^2}{4}=1$

$E_3=\frac{x^2}{9}+\frac{y^2}{25}=1$

Slope of $E_1:\frac{2x}{25}+\frac{2y}{9}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-9x}{25y}$

Slope of $E_2:\frac{dy}{dx}=\frac{-4x}{9y}$

Slope of $E_3:\frac{dy}{dx}=\frac{-25x}{9y}$

To be concyclic points, slopes must be perpendicular/orthoginal intersection of $E_1$ and $E_3$ shows the slopes are otrhogonal.

Thus,intersection of $E_1$ and $E_3$ have 4 concyclic points.