$A point moves as so that the difference of its distances from (ae,0)and(-ae,0)is 2 a, Prove that the equation to its locus is \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, w h e r e b^{2} = a^{2} (e^{2} - 1) .$

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Solution

$Let P(h,k)be any point on the locus and let A(ae,0) and B(-ae,0) be the given points. By the given condition P A - P B = 2 a \Rightarrow P A = 2 a + P B \Rightarrow \sqrt{(a e - h)^{2} + (0 - k)^{2}} = 2 a + \sqrt{(- a e - h)^{2} + (0 - k)^{2}} \Rightarrow {(\sqrt{(a e - h)^{2} + (k)^{2}})}^{2} = {(2 a + \sqrt{(a e - h)^{2} + (k)^{2}})}^{2} [T a k i n g s q u a r e o n b o t h s i d e s] \Rightarrow (a e)^{2} + h^{2} - 2 a e h + k^{2} = 4 a^{2} (a e + h)^{2} + k^{2} + 2 \times 2 a \times \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow h^{2} + k^{2} + (a e)^{2} - 2 a e h = 4 a^{2} + (a e)^{2} + h^{2} + 2 h a e + k^{2} + 4 a \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow - 4 a^{2} - 2 a e h - 2 a e h = 4 a \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow - 4 a^{2} - 4 a e h = 4 a \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow - 4 [a^{2} + a e h] = 4 a \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow - [a^{2} + a e h] = a \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow - a [a + e h] = a \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow - [a + e h] = \sqrt{(a e + h)^{2} + k^{2}} \Rightarrow (a + e h)^{2} = {(\sqrt{(a e + h)^{2} + k^{2}})}^{2} [T a k i n g s q u a r e o n b o t h s i d e s] \Rightarrow a^{2} + (e h)^{2} + 2 h a e = (a e + h)^{2} + k^{2} \Rightarrow a^{2} + (e h)^{2} + 2 h a e = (a e)^{2} + h^{2} + 2 h a e + k^{2} \Rightarrow a^{2} + e^{2} h^{2} = a^{2} e^{2} + h^{2} + k^{2} \Rightarrow e^{2} h^{2} - h^{2} - k^{2} = a^{2} e^{2} - a^{2} \Rightarrow h^{2} (e^{2} - 1) - k^{2} = a^{2} (e^{2} - 1) \Rightarrow \frac{h^{2} (e^{2} - 1)}{^{2} (e^{2} - 1)} - \frac{k^{2}}{a^{2} (e^{2} - 1)} = 1 \Rightarrow \frac{h^{2}}{a^{2}} - \frac{k^{2}}{b^{2}} = 1, w h e r e b^{2} = a^{2} (e^{2} - 1) . ∴ T h e l o c u s o f (h, k) i s \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 H e n c e p r o v e d .$

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