Question

de a and perform the symmetry lest on each of the following is. eld plerd Find the in 16 r2 + 25y2 = 400 A00 (c) r2 + 4y2 = 4 (e) 4x2 + y2 = 64 (B) 9x² + 4y2 = 36 (b) 25x+6y (d) 4x + y = A () Ay? (h) 7x + y - 112 Graph the vertices, foci, endpoints of the minor axis, and endpoints of the latera recta, then draw AB is a chord of the partial ellipse with equation f(x) = ba? - x'. (a) 576x2 + 625y2 = 360,000, A (15,f(15)), B(20, f(20)) (b) 49x 2 + 625y2 = 30,625, A (15, f(15)), B(20,f(20)) – x. Find the length of AB using

Answer 1

The given ellipse is

`576x^2+625y^2=360,000`

Where A(15, f(15)), B(20, f(20)).

First, we find f(15) and f(20) by evaluating the given expression

`\begin{gathered} f(15)=576(15)^2+625y^2=360,000 \\ 576\cdot225+625y^2=360,000 \\ 129,600+625y^2=360,000 \\ 625y^2=360,000-129,600 \\ 625y^2=230,400 \\ y^2=(230,400)/(625) \\ y^2=368.64 \\ y=\sqrt[]{368.64} \\ y=19.2 \end{gathered}`

We use the same process to find f(20).

`\begin{gathered} f(20)=576(20)^2+625y^2=360,000 \\ 576\cdot400+625y^2=360,000 \\ 625y^2=360,000-230,400 \\ x^2=(129,600)/(576) \\ x=\sqrt[]{225}=15 \\ \end{gathered}`

So, the points are A(15, 19.2) and B(20, 15). To find the distance between these points, we have to use the distance formula

`\begin{gathered} d_(AB)=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ d_(AB)=\sqrt[]{(20-15)^2+(15-19.2)^2} \\ d_(AB)=\sqrt[]{5^2+(-4.2)^2}=\sqrt[]{25+17.64}=\sqrt[]{42.64} \\ d_(AB)\approx6.5 \end{gathered}`

Hence, the length of AB is around 6.5 units.