Question

Let $A>B>0$. Show that the distance from any point on the graph of the curve with equation $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$to the point $(-C,0)\text{is}{D}_2=\frac{A^2+Cx}{A^2}\text{, where}C=\sqrt{A^2-B^2}\text{.}$

Short answer

Hence proved.

Step-by-step solution

Step 1. Given information.

Given:$D_2=\frac{A^2+Cx}{A^2},\text{where}C=\sqrt{A^2-B^2}$

Step 2. Explanation.

Now,

$$\begin{gathered} \text{Let}(x,y)\text{be any point on the graph.} \\ \text{Distance between the points}(x,y),(-C,0)\text{is}(x,y),\left(\sqrt{A^2-B^2},0\right)\text{.} \\ \text{Distance}=\sqrt{{\left(x-\left(\sqrt{A^2-B^2}\right)\right)}^2+(y-0{)}^2}\left[since{x}_1=x,y_1=y,x_2=\sqrt{A^2-B^2},y_2=0\right] \\ \text{Distance}=\sqrt{\left(x^2+{\left(\sqrt{A^2-B^2}\right)}^2-2x\sqrt{A^2-B^2}\right)+y^2} \\ =\sqrt{\left(x^2+A^2-B^2-2x\sqrt{A^2-B^2}\right)+y^2} \\ \text{Distance}=\sqrt{\left(x^2+A^2-B^2-2x\sqrt{A^2-B^2}\right)+B^2-\frac{B^2{x}^2}{A^2}} \\ \left[since\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\Rightarrow y^2=B^2-\frac{B^2{x}^2}{A^2}\right] \\ =\sqrt{x^2+A^2-2x\sqrt{A^2-B^2}-\frac{B^2{x}^2}{A^2}} \\ \text{Distance}=\sqrt{\frac{x^2\left(A^2-B^2\right)+A^4-2x{A}^2\sqrt{A^2-B^2}}{A^2}} \\ \text{Distance}=\sqrt{\frac{x^2{c}^2+A^4+2x{A}^{2}c}{A^2}}\left[\text{since}\mathrm{C}=\sqrt{A^2-B^2}\right] \\ \text{Distance}=\frac{\sqrt{{\left(xc+A^2\right)}^2}}{\sqrt{A^2}} \\ \text{Distance}=\frac{A^2+cx}{A} \end{gathered}$$

Hence proved.