Question

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

Answer 1

Step 1: Finding the value of $c$

We know that focus of a parabola is $(a,0)$

Given

$$\begin{gathered} y^2=-64x \\ and4a=-64 \\ \Rightarrow a=-16 \end{gathered}$$

$\therefore \text{Focus}=(-16,0)$

$y=mx+c$ is a focal chord

$0=-16m+c$ [substituting the values]

$c=16m.....\left(1\right)$

Step 2: Finding the value of $m$

The given equation for circle is ${(x+10)}^2+y^2=4$

Also, we know that ${(x-a)}^2+{\left(y-b\right)}^2=r^2$ to find the radius

Hence, the radius $r=2$ and $a=-10,b=0$

$y=mx+c$ is tangent to ${(x+10)}^2+y^2=4$

The formula for perpendicular distance is $\mathbf{d}\mathbf{=}\frac{\left|{\mathrm{ax}}_1+{\mathrm{by}}_1+c\right|}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}}$

Substituting the values we get

$$\begin{gathered} 2=\frac{\left|-16m+10m\right|}{\sqrt{1+m^2}}\mathbf{[}\mathbf{\because }\mathbf{-}\mathbf{a}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathbf{c}\mathbf{=}\mathbf{16}\mathbf{m}\mathbf{]} \\ \Rightarrow 2\sqrt{1+m^2}=\pm 6m \\ \Rightarrow 4(1+m^2)=36m^2 \\ \Rightarrow 4+4m^2=36m^2 \\ \Rightarrow 32m^2=4 \\ \Rightarrow m^2=\frac{4}{32} \\ \Rightarrow m=\frac{1}{2\sqrt{2}}\left(m>0\right) \end{gathered}$$

Step 3: Finding the value of $4\sqrt{2}(m+c)$

Therefore,

$\begin{array}{rcl}4\sqrt{2}(m+c)& =& 4\sqrt{2}\left(16m+m\right)\left[\because c=16m\right]\\ & =& 4\sqrt{2}\times 17m\\ & =& 4\sqrt{2}\times 17\frac{1}{2\sqrt{2}}\\ & =& 34\end{array}$

Hence, the value of $4\sqrt{2}\left(m+c\right)=34$