Question

A chord of negative slope is drawn from the point $P(\sqrt{264},0)$ to $x^2+4y^2=16.$ If the chord intersects the curve at $A$ and $B,$ where $O$ is the origin, then which of the following is (are) CORRECT ?

• A. The maximum area of $△AOB$ is $4$ sq. units
• B. The maximum area of $△AOB$ is $8$ sq. units
• C. When the area of $△AOB$ is maximum, then slope of line $AB$ is $\frac{-1}{2\sqrt{2}}$
• D. When the area of $△AOB$ is maximum, then slope of line $AB$ is $\frac{-1}{8\sqrt{2}}$

Answer 1

Answer: (A), (D)

When the area of $△AOB$ is maximum, then slope of line $AB$ is $\frac{-1}{8\sqrt{2}}$

Given : $x^2+4y^2=16$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{4}=1$
Let point $A=(4\cos {\theta }_1,2\sin {\theta }_1)$ and $B=(4\cos {\theta }_2,2\sin {\theta }_2)$


Equation of chord $AB$ in parametric form is
$\frac{x}{4}\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)+\frac{y}{2}\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)$
This chord passes through $P(\sqrt{264},0)$, so
$\sqrt{\frac{33}{2}}\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)\cdots \left(1\right)$

Now, the area of $△AOB$
$=\frac{1}{2}\left|\begin{array}{ccc}4\cos {\theta }_1& 2\sin {\theta }_1& 1\\ 4\cos {\theta }_2& 2\sin {\theta }_2& 1\\ 0& 0& 1\end{array}\right|=4|\sin ({\theta }_2-{\theta }_1)|$

So, maximum area $=4$ sq. units
The triangle area is maximum when
$|\sin ({\theta }_2-{\theta }_1)|=1\Rightarrow ({\theta }_2-{\theta }_1)=\frac{(2n+1)\pi }{2},n\in Z$
From equation $\left(1\right)$, we get
$\sqrt{\frac{33}{2}}\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\pm \frac{1}{\sqrt{2}}\Rightarrow \cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\pm \frac{1}{\sqrt{33}}$
Slope of line $AB$ is
$m=-\frac{1}{2}\cot \left(\frac{{\theta }_1+{\theta }_2}{2}\right)\Rightarrow m=\mp \frac{1}{2}\times \frac{1}{\sqrt{32}}=\pm \frac{1}{8\sqrt{2}}$
As $AB$ should be negative, so
$m=-\frac{1}{8\sqrt{2}}$