Question

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

• A. $\left\{–1,1\right\}$
• B. $\left\{–2,2\right\}$
• C. $\left\{–2,-\frac{1}{2}\right\}$
• D. $\left\{2,-\frac{1}{2}\right\}$

Answer 1

The correct option is A

$\left\{–1,1\right\}$

Explanation for the correct option:

Step-1 Length of tangent :

Given: The focal chord to $y^2=16x$ is tangent to ${(x–6)}^2+y^2=2$

The standard equation of the parabola is:$y^2=4ax$

Comparing the given equation with the above equation we get, $a=4$

Therefore the focus of the parabola is$(4,0).$

This means that the tangents are drawn from $(4,0)$ to ${(x–6)}^2+y^2=2$

$PA$ is the tangent to the circle as shown in figure.

As radius is perpendicular to the tangent, we have length of tangent from $(4,0)$ to the circle is = $\sqrt{S_1}$

$\Rightarrow AP=\sqrt{{(4-6)}^2+0-2}=\sqrt{2}$

Step-2 Slope of tangents:

General equation of circle

${(x–a)}^2+{(y-b)}^2=r^2$, where $(\mathrm{a},\mathrm{b})$ is centre and $r$ is radius.

Here, Focal chord of the parabola is tangent to the circle ${(x–6)}^2+y^2=2$ so the $\sqrt{2}$ and $\left(6,0\right)$ are the radius and centre of the circle.

We know that,

$\tan \theta =$slope of tangent $=\frac{\mathrm{AC}}{AP}=\frac{\sqrt{2}}{\sqrt{2}}=1$

So, $\frac{\mathrm{BC}}{\mathrm{BP}}=-1$

Thus, the slope of the focal chord as the tangent to the circle is $\pm 1$.

Hence, the possible values of the slope of this chord are $\left\{–1,1\right\}$.

Therefore, option (A) is the correct answer.