Question

A point moves such that its distance from the point $(4,0)$ is half that of its distance from the line $x=16$.

The locus of this point is

• A. $3x^2+4y^2=192$
• B. $4x^2+3y^2=192$
• C. $x^2+y^2=192$
• D. None of these

Answer 1

The correct option is A

$3x^2+4y^2=192$

Finding the locus of the point:

Let, $P(h,k)$ be the point and the given point is $A(4,0)$

Distance of the point $(x_1,y_1)$from the line $ax+by+c=0$is

$D=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$D=\left|\frac{h-16}{1^2+0^2}\right|$here, $a=h,x=1,y=0,b=k$

$AP=\sqrt{{(h-4)}^2+k^2}$

$\Rightarrow \sqrt{{(h-4)}^2+k^2}=\frac{1}{2}\left|\frac{h-16}{1^2+0^2}\right|$

Squaring both side

$\begin{array}{rcl}{(h-4)}^2+k^2& =& {\left(\frac{1}{2}\left|\frac{h-16}{1^2+0^2}\right|\right)}^2\\ h^2-8h+16+k^2& =& \frac{{(h-16)}^2}{4}\\ 4h^2-32h+64+4k^2& =& h^2-32h+256\\ 3h^2+4k^2& =& 192\end{array}$

So the locus of point is

$3x^2+4y^2=192$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{.}$