Question

The focal chord to $y^2=16x$ is tangent to ${(x-6)}^2+y^2=2$, then the possible value of the slope of this chord are

• A. $(-1,1)$
• B. $(-2,2)$
• C. $(-2,\frac{1}{2})$
• D. $(2,-\frac{1}{2})$

Answer 1

Answer: (A)

$(-1,1)$

$y^2=16x$ .....(i)

Here, $4a=16\Rightarrow a=4$

Focus of the parabola (i) is $(4,0)$

Focal chord of the parabola is tangent to the circle ${(x-6)}^2+y^2=2$.

$\sqrt{2}$ and $(6,0)$ are radius and centre of the circle

As radius is perpendicular to the tangent, we have length of tangent from $(4,0)$ to the circle is $=\sqrt{2}$.

From the diagram, we have
$\tan \theta =\frac{\sqrt{2}}{\sqrt{2}}=1\Rightarrow \theta ={45}^o$
Therefore, slope of the chord is $\pm 1=(-1,1)$.