Chapter 9: Q. 55 (page 773)

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 $(- C, 0) is D_{2} = \frac{A^{2} + C x}{A^{2}}, where C = \sqrt{A^{2} - B^{2}} .$

Short Answer

Expert verified

Hence proved.

Step by step solution

Step 1. Given information.

Given: $D_{2} = \frac{A^{2} + C x}{A^{2}}, where C = \sqrt{A^{2} - B^{2}}$

Step 2. Explanation.

Now,

$Let (x, y) be any point on the graph. Distance between the points (x, y), (- C, 0) is (x, y), (\sqrt{A^{2} - B^{2}}, 0) . Distance = \sqrt{{(x - (\sqrt{A^{2} - B^{2}}))}^{2} + (y - 0)^{2}} [since x_{1} = x, y_{1} = y, x_{2} = \sqrt{A^{2} - B^{2}}, y_{2} = 0] Distance = \sqrt{(x^{2} + {(\sqrt{A^{2} - B^{2}})}^{2} - 2 x \sqrt{A^{2} - B^{2}}) + y^{2}} = \sqrt{(x^{2} + A^{2} - B^{2} - 2 x \sqrt{A^{2} - B^{2}}) + y^{2}} Distance = \sqrt{(x^{2} + A^{2} - B^{2} - 2 x \sqrt{A^{2} - B^{2}}) + B^{2} - \frac{B^{2} x^{2}}{A^{2}}} [since \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1 \Rightarrow y^{2} = B^{2} - \frac{B^{2} x^{2}}{A^{2}}] = \sqrt{x^{2} + A^{2} - 2 x \sqrt{A^{2} - B^{2}} - \frac{B^{2} x^{2}}{A^{2}}} Distance = \sqrt{\frac{x^{2} (A^{2} - B^{2}) + A^{4} - 2 x A^{2} \sqrt{A^{2} - B^{2}}}{A^{2}}} Distance = \sqrt{\frac{x^{2} c^{2} + A^{4} + 2 x A^{2} c}{A^{2}}} [since C = \sqrt{A^{2} - B^{2}}] Distance = \frac{\sqrt{{(x c + A^{2})}^{2}}}{\sqrt{A^{2}}} Distance = \frac{A^{2} + c x}{A}$

Hence proved.

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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 $(- C, 0) is D_{2} = \frac{A^{2} + C x}{A^{2}}, where C = \sqrt{A^{2} - B^{2}} .$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Explanation.

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Let A>B>0. Show that the distance from any point on the graph of the curve with equation x2A2+y2B2=1to the point (-C,0)isD2=A2+CxA2, whereC=A2-B2.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Explanation.

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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 $(- C, 0) is D_{2} = \frac{A^{2} + C x}{A^{2}}, where C = \sqrt{A^{2} - B^{2}} .$