Question

Focal chord to $y^2=16xistangentto(x-6{)}^2+y^2=2$ then the possible values of the slopes of this chord(s),are

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Here we first draw the parabola and given circle of radius $\sqrt{2}$ with centre (6, 0). The focal chord passes, through focus (4, 0) there are 2 possible ways in which focal chord can touch the given circle.

In both cases the perpendicular from centre of the circle to the focal chord will be equal to the radius of the circle.

Lets take $△APRand△AQR$

$PR=\sqrt{2};AR=2$

$\Rightarrow AP=\sqrt{2}$

$\therefore \angle PAR=si{n}^{-1}\left(\frac{\sqrt{2}}{2}\right)=si{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$

$\therefore slopeofAP=tan\frac{\pi }{4}=1$

Slope of $AQ=tan(-\frac{\pi }{4})=-1$

$\therefore$ Range of slope={-1,1}