Question

the graph of the ellipse is given.
(a) use the distance formula to find the distances ( d_1 ), ( d_2 ), ( d_3 ), and ( d_4 ).
(b) find the sum ( d_1 + d_2 ).
(c) find the sum ( d_3 + d_4 ).
(d) how do the sums from (b) and (c) compare?
(e) how do the sums of the distance from (b) and (c) relate to the length of the major axis?
express numbers in exact, simplest form. note that the points ( (-4,0) ) and ( (4,0) ) represent the foci.

Explanation:

Step1: Use distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(d_1\): Points are \((- 4,0)\) and \((0,-3)\).
\(d_1=\sqrt{(0 + 4)^2+(-3-0)^2}=\sqrt{16 + 9}=\sqrt{25}=5\).
For \(d_2\): Points are \((4,0)\) and \((0,-3)\).
\(d_2=\sqrt{(0 - 4)^2+(-3-0)^2}=\sqrt{16 + 9}=\sqrt{25}=5\).

Step2: Calculate \(d_1 + d_2\)

\(d_1 + d_2=5 + 5 = 10\).
For \(d_3\): Points are \((-4,0)\) and \((3,\frac{12}{5})\).
\(d_3=\sqrt{(3 + 4)^2+(\frac{12}{5}-0)^2}=\sqrt{49+\frac{144}{25}}=\sqrt{\frac{1225+144}{25}}=\sqrt{\frac{1369}{25}}=\frac{37}{5}\).
For \(d_4\): Points are \((4,0)\) and \((3,\frac{12}{5})\).
\(d_4=\sqrt{(3 - 4)^2+(\frac{12}{5}-0)^2}=\sqrt{1+\frac{144}{25}}=\sqrt{\frac{25 + 144}{25}}=\sqrt{\frac{169}{25}}=\frac{13}{5}\).

Step3: Calculate \(d_3 + d_4\)

\(d_3 + d_4=\frac{37}{5}+\frac{13}{5}=\frac{37 + 13}{5}=\frac{50}{5}=10\).

Answer:

(b) \(d_1 + d_2 = 10\). (c) \(d_3 + d_4=10\). (d) The sums from (b) and (c) are equal. (e) The sum of the distances from any point on an ellipse to the two foci is equal to the length of the major axis.