First slide

Parabola

Question

$The focal chord of the parabola (y - 2)^{2} = 16 (x - 1) is a tangent to the circle$
$x^{2} + y^{2} - 14 x - 4 y + 51 = 0, then slope of the focal chord can be$

Moderate

A

0
B

1
C

2
D

3

Solution

$Given parabola is (y - 2)^{2} = 16 (x - 1)$
$\begin{array}{l} Here vertex = (h, k) = (1, 2) \\ 4 a = 16 \Rightarrow a = 4 \\ Focus = (h + a, k) = (1 + 4, 2) = (5, 2) \\ The equation of focal chord is y - 2 = m (x - 5) \\ \Rightarrow m x - y + 2 - 5 m = 0 \dots \dots \dots . . . (1) \end{array}$
$Given circle equation is x^{2} + y^{2} - 14 x - 4 y + 51 = 0 \dots \dots \dots . . . (2)$
$\begin{array}{l} Hence centre (7, 2) \\ And radius, r = \sqrt{49 + 4 - 51} = \sqrt{2} \end{array}$
Since equation (1) is a tangent of circle equation (2) then
r = d
$\begin{array}{l} \sqrt{2} = \frac{| 7 m - 2 + 2 - 5 m |}{\sqrt{m^{2} + 1}} \\ (∵ d = perpendiculardistance from centre to equation (1)) \end{array}$
$\begin{aligned} \Rightarrow \sqrt{2} \sqrt{m^{2} + 1} = | 2 m | \\ Squaring on both sides \end{aligned}$
$\begin{array}{l} \Rightarrow 2 m^{2} = m^{2} + 1 \\ \Rightarrow m = \pm 1 \\ ∴ Slope of focal chord is 1 \end{array}$

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