Question

If a parabola whose length of latus rectum is $4$a touches both the coordinate axes then the locus of its focus is?

• A. $xy=a^2(x^2+y^2)$
• B. $x^2{y}^2=a^2(x^2+y^2)$
• C. $x^2-y^2=a^2(x^2+y^2)$
• D. $x^2{y}^2=a^2(x^2-y^2)$

Answer 1

The correct option is C $x^2-y^2=a^2(x^2+y^2)$

Parabolo has equation $y^2=4ax$ with latus rectum $4a$.

If $x=a{t}^2,y=2at$ is any point on this parabola

slope $=\left(\frac{dy}{dt}\right)\left(\frac{dx}{dt}\right)=\frac{2a}{2at}=\frac{1}{t}$ ----- $\left(i\right)$

So, the tanpent at $t$ is $y-2at=\left(\frac{1}{t}\right)(x-a{t}^2)$

$\Rightarrow x-ty+a{t}^2=0$ ---- (ii)$

If $2$ tangent at $r$ & $s$ are mutually perpendicular then from eq$\left(1\right)\Rightarrow \left(\frac{1}{r}\right)\left(\frac{1}{5}\right)=-1$ or $rs=-1$ ---- $\left(iii\right)$

Let distance of focus $(a,0)$ from there tangent be $x$ & $y$.

By point distance formula applied to $\left(ii\right)$

$x=\frac{|a-r\times 0+a{r}^2|}{\sqrt{1+r^2}}=asqrt1+r^2$ --- $\left(iv\right)$

Similarly $y=a\sqrt{1+5^2}$ ---- $\left(v\right)$

from eq $\left(iv\right)$ & $\left(v\right)$, $r^2=\frac{x^2}{a^2}-1$ & $s^2=\frac{y^2}{a^2}-1$

But from eq $\left(iii\right)\Rightarrow r^2{s}^2=1$

So, $(\frac{x^2}{a^2}-1)(\frac{y^2}{a^2}-1)=1$

$(x^2-a^2)(y^2-a^2)=a^4$

$x^2{y}^2-a^2(x^2+y^2)+t{a}^4=a^4$

$x^2{y}^2=a^2(x^2+y^2)$ ---- $\left(vi\right)$

Now consider the situation when this same parabola is moved so that it touches the $xy$ axis.

Axis are pair of perpendicular tangent a relationship like $\left(vi\right)$ will still hold. However in this case $x,y$ are the coordinates of focus.

So, equation $\left(vi\right)$ is deserved locus