Question

If point $(4,4)$ lies on a parabola whose focus lies on $x$ -axis and directrix is $lx+my=1$ such that $(l,m)$ lies on curve $x^2+y^2-32xy+8x+8y-1=0,$ then

• A. Co-ordinates of focus may be $(5,0)$
• B. Co-ordinates of focus may be $(3,0)$
• C. If tangent at $(4,4)$ passes through origin, value of $m$ may be $\frac{1}{3}$
• D. If tangent at $(4,4)$ passes through origin, equation of axis may be $x-4y-3=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Co-ordinates of focus may be $(5,0)$
B Co-ordinates of focus may be $(3,0)$
C If tangent at $(4,4)$ passes through origin, value of $m$ may be $\frac{1}{3}$
D If tangent at $(4,4)$ passes through origin, equation of axis may be $x-4y-3=0$

Let focus of the parabola is $(\alpha ,0)$
Now, $SP=PM$
$\Rightarrow \sqrt{(4-\alpha {)}^2+(4-0{)}^2}=\frac{|4l+4m-1|}{\sqrt{l^2+m^2}}$



$\Rightarrow (l^2+m^2)\left(\right(\alpha -4{)}^2+16)=\left(4\right(l+m)-1{)}^2$
$\Rightarrow (l^2+m^2)({\alpha }^2+32-8\alpha )=16(l+m{)}^2+1-8(l+m)$
$\Rightarrow (l^2+m^2)({\alpha }^2+32-8a-16)+8(l+m)-32lm=1$
$\Rightarrow (l^2+m^2)(\alpha -4{)}^2+8(l+m)-32lm-1=0$

But given that $(l^2+m^2)+8(l+m)-32lm-1=0$
$\Rightarrow (\alpha -4{)}^2=1$
$\therefore \alpha =5$ or $3$.

Take focus $(3,0)$
Equation of directrix $lx+my=1$ satisfies $(0,3)$ i.e. mirror image of focus through tangent $y=x$
$\Rightarrow 0+3m=1$
$\Rightarrow m=\frac{1}{3}$

Slope of line joining $(0,3)$ and $(4,4)$ is $\frac{1}{4}$ which is parallel to axis.
$\therefore$ Equation of axis $y-0=\frac{1}{4}(x-3)$
$\Rightarrow x-4y-3=0$