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# Help!

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Let $$F_1=(10,2)$$and $$F_2 = (-16,2)$$ . Then the set of points $$P$$ such that $$|PF_1 - PF_2| = 24$$ form a hyperbola. The equation of this hyperbola can be written as $$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$. Find $$h + k + a + b.$$

Aug 4, 2019

#1
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Let $$F_1=(10,2)$$ and  $$F_2 = (-16,2)$$. Then the set of points $$P$$ such that $$|PF_1 - PF_2| = 24$$ form a hyperbola.
The equation of this hyperbola can be written as $$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$$.
Find $$h + k + a + b$$.

$$\begin{array}{|rcll|} \hline \mathbf{|PF_1 - PF_2|} &=& \mathbf{2a} \\ |PF_1 - PF_2| &=& 24 \\ 2a&=& 24 \\ \mathbf{a} &=& \mathbf{12} \\ \hline \end{array}$$

$$\begin{array}{|lrc|} \hline \mathbf{\text{Focus}_1:}& \mathbf{F_1(h+ae,k)} \\ & F_1(10,2): & h+ae = 10 \quad (1) \\ && \mathbf{ k =2} \\ \hline \mathbf{\text{Focus}_2:}& \mathbf{F_2(h-ae,k)} \\ & F_2(-16,2): & h-ae = -16\quad (2) \\ && \mathbf{ k =2} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1)+(2): & h+ae + h-ae &=& 10-16 \\ & 2h &=& -6 \\ & \mathbf{ h }&=& \mathbf{-3} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1) & h+ae &=& 10 \quad |\quad h=-3 \\ & -3+ae &=& 10 \\ & \mathbf{ ae }&=& \mathbf{13} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline a^2 + b^2 &=& (ae)^2 \\ 12^2+b^2 &=& 13^2 \\ b^2 &=& 13^2-12^2 \\ b^2 &=& 169-144 \\ b^2 &=& 25 \\ \mathbf{ b }&=& \mathbf{5} \\ \hline \end{array}$$

$$\begin{array}{rcll} && \mathbf{h + k + a + b} \\ &=& -3 + 2 +12+5 \\ &=& \mathbf{16} \\ \end{array}$$

Aug 5, 2019