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Question Number 99159 by bemath last updated on 19/Jun/20

Commented by som(math1967) last updated on 19/Jun/20

$$\mathrm{let}\mathrm{pt}.\mathrm{on}\mathrm{paabola}(\mathrm{h},\mathrm{k})$$$$\therefore \sqrt{{(\mathrm{h}+3)}^2+{(\mathrm{k}-3)}^2}=\frac{\mid \mathrm{k}-7\mid }{\sqrt{1^2}}$$$$\Rightarrow {(\mathrm{h}+3)}^2+{(\mathrm{k}-3)}^2={(\mathrm{k}-7)}^2$$$$\mathrm{putting}(\mathrm{x},\mathrm{y})\mathrm{in}\mathrm{the}\mathrm{eqn}.$$$$\Rightarrow {(\mathrm{x}+3)}^2={(\mathrm{y}-7)}^2-{(\mathrm{y}-3)}^2$$$${\mathrm{x}}^2+6\mathrm{x}+9=2(\mathrm{y}-5)(-4)$$

Commented by bemath last updated on 19/Jun/20

$$\mathrm{yes}.\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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