Q. 55 - Calculus

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 $(-C,0)\text{ is }{D}_2=\frac{A^2+Cx}{A^2}\text{, where }C=\sqrt{A^2-B^2}\text{. }$

Step-by-Step Solution

Answer

Hence proved.

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.