Question

For all real $p\in [-1,1],$ the line $2px+y\sqrt{1-p^2}=1$ touches a fixed ellipse whose axes are coordinate axes.
Then which of the following is/are true

• A. Foci of the ellipse are $(0,\pm \frac{1}{2})$
• B. Foci of the ellipse are $(0,\pm \frac{\sqrt{3}}{2})$
• C. Director circle of the ellipse is $x^2+y^2=\frac{3}{4}$
• D. Director circle of the ellipse is $x^2+y^2=\frac{5}{4}$

Answer 1

Answer: (B), (D)

The correct options are
B Foci of the ellipse are $(0,\pm \frac{\sqrt{3}}{2})$
D Director circle of the ellipse is $x^2+y^2=\frac{5}{4}$

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Equation of tangent to ellipse at $(x_1,y_1)$ will be
$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

Hence, it it identical with
$2px+y\sqrt{1-p^2}=1$
$\Rightarrow \frac{x_1}{a^2}=2p,\frac{y_1}{b^2}=\sqrt{1-p^2}$

Hence from ellipse equation
$p^2(4a^2-b^2)+b^2-1=0$

This equation is true for all real $p\in [-1,1]$, if
$b^2=1$ and $4a^2=b^2$
$\Rightarrow b^2=1$ and $a^2=\frac{1}{4}$

Therefore, the equation of the ellipse is
$\frac{4x^2}{1}+\frac{y^2}{1}=1$

Now $a^2=b^2(1-e^2)$
$\Rightarrow \frac{1}{4}=1-e^2\Rightarrow e=\frac{\sqrt{3}}{2}$
hence foci are $(0,\pm \frac{\sqrt{3}}{2})$

Equation of director circle is
$x^2+y^2=b^2+a^2$
$\Rightarrow x^2+y^2=\frac{5}{4}.$