Question

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

• A. $\pm 1$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $2$

Answer 1

The correct option is A $\pm 1$

$y^2=16x$

$\text{focus}=(4,0)$

$\text{Equation of line passing through}(4,0)\text{whose slope}$

$\text{is}m\text{(say) is given by}$

$y-0=m(x-4)$

$\Rightarrow y=mx-4m-\left(4\right)-\left(L1\right)$

$\text{Now, equation of given circle is}(x-6{)}^2+y^2=2$

$\therefore \text{Distance of line}4\text{from}(6,0)\text{nill be given by}$

$\left|\frac{y_1-m{x}_1+4m}{\sqrt{1+m^2}}\right|=\sqrt{2}$

$\Rightarrow \frac{\mid 0-6m+4m\mid }{\sqrt{1+m^2}}=\sqrt{2}$

$\text{Squaring both sides and solving we get}$

$4m^2=2(1+m^2)$

$2m^2=2$

$m=\pm 1$

Hence, option A is correct.