Question

If the base $BC$ of a $△ABC$ passes through $(4,8)$ and its other two sides are bisected at right angle by the pair of lines $x^2-9y^2-8xy=0$, then which of the following is (are) CORRECT?

• A. The locus of the point $A$ is a circle.
• B. The locus of the point $A$ is an ellipse.
• C. $\angle BAC=\pi -{\tan }^{-1}\frac{5}{4}$
• D. $\angle BAC=\pi -{\tan }^{-1}\frac{4}{5}$

Answer 1

Answer: (A), (C)

The correct options are
A The locus of the point $A$ is a circle.
C $\angle BAC=\pi -{\tan }^{-1}\frac{5}{4}$

$x^2-9y^2-8xy=0$
Acute angle between the pair of lines is,
$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|\Rightarrow \tan \theta =\left|\frac{2\sqrt{4^2+9}}{-8}\right|\Rightarrow \tan \theta =\frac{5}{4}\therefore \angle BAC=\pi -{\tan }^{-1}\frac{5}{4}[\because \angle APO=\angle AQO=\frac{\pi }{2}]$

Now, $x^2-9y^2-8xy=0$
$\Rightarrow (x+y)(x-9y)=0\Rightarrow x+y=0,x-9y=0$
Assuming Equation of line
$OP:x+y=0OQ:x-9y=0$
Let the coordinate of $A=(h,k)$

Coordinates of $B:$
Reflection of point $A$ about the line $OP:x+y=0$
$\frac{x-h}{1}=\frac{y-k}{1}=-\frac{2(h+k)}{1^2+1^2}\Rightarrow x=-k,y=-hB=(-k,-h)$

Coordinates of $C:$
Reflection of point $A$ about the line $OQ:x-9y=0$
$\frac{x-h}{1}=\frac{y-k}{-9}=-\frac{2(h-9k)}{1^2+9^2}\Rightarrow x=\frac{40h+9k}{41},y=\frac{9h-40k}{41}C=(\frac{40h+9k}{41},\frac{9h-40k}{41})$
It is given that the base passes through $(4,8)$.
So, $\frac{8+h}{4+k}=\frac{\frac{9h-40k}{41}+h}{\frac{40h+9k}{41}+k}\Rightarrow \frac{8+h}{4+k}=\frac{\frac{50h-40k}{41}}{\frac{40h+50k}{41}}\Rightarrow \frac{8+h}{4+k}=\frac{5h-4k}{4h+5k}\Rightarrow 32h+40k+4h^2+5hk=20h-16k+5hk-4k^2\Rightarrow 4h^4+4k^2+12h+56k=0\Rightarrow h^2+k^2+3h+14k=0$
Therefore, the locus of vertex $A$ is,
$x^2+y^2+3x+14y=0$
Hence, the locus of $A$ is circle.