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Let y=m x+c, m>0 be the focal chord of \mathrm{y^2=-64 x}, which is tangent to \mathrm{(x+10)^2+y^2=4.} Then the value of \mathrm{2 \sqrt{2}(m+c)} is equal to

Option: 1

17


Option: 2

18


Option: 3

19


Option: 4

20


Answers (1)

best_answer

\mathrm{\text { We have, } y^2=-64 x}

Focus = (–16, 0).
Now, y = mx + c is the focal chord.
⇒ c = 16m                ...[i]

\mathrm{\text { Also, } y=m x+c \text { is tangent to }(x+10)^2+y^2=4}

\mathrm{\begin{aligned} & \therefore \quad y=m(x+10) \pm 2 \sqrt{1+m^2} \\ & \Rightarrow c=10 m \pm 2 \sqrt{1+m^2} \end{aligned}}           ....[ii]

From (i) and (ii), we get

\mathrm{\begin{aligned} & 16 m=10 m \pm 2 \sqrt{1+m^2} \\ & \Rightarrow 6 m=2 \sqrt{1+m^2} \\ & \Rightarrow 9 m^2=1+m^2 \\ & \Rightarrow m=\frac{1}{2 \sqrt{2}} \text { and } c=\frac{8}{\sqrt{2}} \\ & \therefore \quad 2 \sqrt{2}(m+c)=2 \sqrt{2}\left(\frac{17}{2 \sqrt{2}}\right)=17 \end{aligned}}

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