The focal chord to y 2 = 16 x is tangent to ( x - 6 ) 2 + y 2 = 2 , then the possible values of the slope of this chord are [2003]

Question

The focal chord to $y^{2} = 16 x$ is tangent to ${(x - 6)}^{2} + y^{2} = 2$ , then the possible values of the slope of this chord are [2003]

Smart Achievers · Accepted Answer

(1)

Given parabola $y^{2} = 16 x$ , its focus = (4, 0). Let $m$ be the slope of focal chord then its equation is

$y = m (x - 4) ...(i)$

But it is given that equation (i) is a tangent to the circle

${(x - 6)}^{2} + y^{2} = 2$ with centre, C(6, 0) and radius $r = \sqrt{2}$

$∴$ Length of perpendicular from (6, 0) to (i) $= \sqrt{2}$

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

$\Rightarrow 2 m^{2} = m^{2} + 1$ $\Rightarrow m = \pm 1$

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