Question

Straight line $12x+5y-60=0$ intersects the curve $144x^2+25y^2-3600=0$ at pts. $P$ and $Q$.
1. There lies a point ${}^{\prime }{T}^{\prime }$ on the $2^{nd}$ curve, such that area of $△TPQ$ is $13sq.units$.
2. Then select the proper options.

• A. $T$ lies on Origin side of line $PQ$
• B. $T$ lies on Non-Origin side of line $PQ$
• C. $T$ does not lie in First Quadrant
• D. $T$ lies in First Quadrant

Answer 1

Answer: (A), (C)

$T$ does not lie in First Quadrant
$T$ lies on Origin side of line $PQ$

Given line is $\frac{x}{5}+\frac{y}{12}=1$
Given curve is $\frac{x^2}{25}+\frac{y^2}{144}=1$ (an Ellipse)
Now, Point $T$ can either lie on Origin side of line $PQ$
OR
On Non-Origin side of line $PQ$ (ie in Quadrant $1$)

Suppose $T$ lies on Non-Origin side of line $PQ$. (AS SHOWN IN FIGURE)
$\Rightarrow T(5\cos \theta ,12\sin \theta )$ and $0<\theta <\pi /2$
Now, $Ar\left(△TPQ\right)=\frac{1}{2}\times PQ\times TM$
Where $T<=\perp r$ distance ${}^{\prime }{d}^{\prime }$ of point $T$ from line $PQ$
$TM=d=\frac{1}{\sqrt{{12}^2+5^2}}|12\left(5\cos \theta \right)+5\left(12\cos \theta \right)-60|$
$d=\frac{60}{13}|\cos \theta +\sin \theta -1|$
Let $f\left(\theta \right)=\cos \theta +\sin \theta$
$f^{\prime }\left(\theta \right)=-\sin \theta +\cos \theta$
$f^{\prime \prime }\left(\theta \right)=-\cos \theta -\sin \theta$
$f^{\prime }\left(\theta \right)=0\Rightarrow \theta =\pi /4$
$f^{\prime \prime }\left(\pi /4\right)=-ve$
$\Rightarrow f\left(\theta \right)$ has a maximum at $\theta =\pi /4$
So, $d_{max}=\frac{60}{13}(\cos \frac{\pi }{4}+\sin \frac{\pi }{4}-1)$
$d_{max}=\frac{60}{13}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)$
$d_{max}=\frac{60}{13}(\sqrt{2}-1)$
So $d\le d_{max}$ ie $d\le \frac{60}{13}(\sqrt{2}-1)$
$Area\left(△TPQ\right)=\frac{1}{2}\times PQ\times TM$
$=\frac{1}{2}\times PQ\times d$
$Ar\left(△TPQ\right)=\frac{1}{2}\times 13\times d\le \frac{1}{2}\times 13\times d_{max}$
$Ar\left(△TPQ\right)\le \frac{1}{2}\times 13\times \frac{60}{13}(\sqrt{2}-1)$
$Ar\left(△TPQ\right)\le 30(\sqrt{2}-1)$
$Ar\left(△TPQ\right)\le 12.42\left(approx\right)$.
But, we are given $Area\left(△TPQ\right)=13sq.units$
So, $T$ cannot lie on Non-Origin side
ie $T$ cannot lie in Quadrant $1$.
$\Rightarrow T$ lies on Origin side of line $PQ$.

SO OUR ASSUMPTION THAT $T$ lies on Non-Origin side of $PQ$ is WRONG.

Hence, our illustrated figure is also wrong, for the given conditions.
Select options $\left(A\right)$ and $\left(C\right)$.