Question

Circles drawn on the diameter as a focal distance of any point lying on the parabola $x^2-4x+6y+10=0$ will touch a fixed line whose equation is:

• A. $y=\frac{1}{2}$
  
• B. $y=-1$
• C. $x+y=2$
• D. $x-y=2$

Answer 1

The correct option is A $y=\frac{1}{2}$

REF.Image

$x^2-4x+6y+10=0$

$\Rightarrow (x-2{)}^2=-4.\frac{3}{2}(y+1)$

$F\equiv (2,(-1-\frac{3}{2}\left)\right)$ [Focus]

$=(2,\frac{-5}{2})$

$C\to$ any point on the parabola = $(h,k)$

Slope of $FC=\left(\frac{k+\frac{5}{2}}{n-2}\right)$

Slope of the line touching the circle.

${}_{(h,k)}\frac{d}{dx}(x^2-4y+6y+10)$

${\frac{dy}{dx}}_{(h,k)}=\frac{2-h}{3}$

$\because$ FC is $\perp$ to the line.

$\left(\frac{2-h}{3}\right)\left(\frac{h+\frac{5}{2}}{h-2}\right)=-1$

$\Rightarrow \frac{K+\frac{5}{2}}{3}=1$

$\Rightarrow k=3-\frac{5}{2}$

$\Rightarrow k=\frac{1}{2}$

$\therefore$ the line is $y=\frac{1}{2}$