The locus of a point whose difference of dist | vedclass

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Question

(A) Let the point be $P(h, k)$.Given that $|PA - PB| = 4$.Let $A = (3, 0)$ and $B = (-3, 0)$.$\sqrt{(h - 3)^2 + k^2} - \sqrt{(h + 3)^2 + k^2} = \pm 4$

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$\frac{x^2}{4} - \frac{y^2}{5} = 1$

Vedclass · Answer

(A) Let the point be $P(h, k)$.Given that $|PA - PB| = 4$.Let $A = (3, 0)$ and $B = (-3, 0)$.$\sqrt{(h - 3)^2 + k^2} - \sqrt{(h + 3)^2 + k^2} = \pm 4$.Squaring both sides:$(h - 3)^2 + k^2 = 16 + (h + 3)^2 + k^2 \mp 8\sqrt{(h + 3)^2 + k^2}$.$h^2 - 6h + 9 + k^2 = 16 + h^2 + 6h + 9 + k^2 \mp 8\sqrt{(h + 3)^2 + k^2}$.$-12h - 16 = \mp 8\sqrt{(h + 3)^2 + k^2}$.Dividing by $-4$: $3h + 4 = \pm 2\sqrt{(h + 3)^2 + k^2}$.Squaring again:$9h^2 + 24h + 16 = 4(h^2 + 6h + 9 + k^2)$.$9h^2 + 24h + 16 = 4h^2 + 24h + 36 + 4k^2$.$5h^2 - 4k^2 = 20$.Dividing by $20$: $\frac{h^2}{4} - \frac{k^2}{5} = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{4} - \frac{y^2}{5} = 1$.

The locus of a point whose difference of distance from points $(3, 0)$ and $(-3, 0)$ is $4$,is

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