A-line-passes-through-A-3-0-and-B-0-4-A-variable-line-perpendicular-to-AB-is-drawn-to-cut-x-and-y-axes-at-M-and-N-Find-the-locus-of-the-point-of-intersection-of-the-lines-AN-and-BM-

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Question Number 25930 by Tinkutara last updated on 16/Dec/17

$$\mathrm{A}\mathrm{line}\mathrm{passes}\mathrm{through}A(-3,0)\mathrm{and}$$$$B(0,-4).\mathrm{A}\mathrm{variable}\mathrm{line}\mathrm{perpendicular}$$$$\mathrm{to}AB\mathrm{is}\mathrm{drawn}\mathrm{to}\mathrm{cut}x\mathrm{and}y-\mathrm{axes}\mathrm{at}$$$$M\mathrm{and}N.\mathrm{Find}\mathrm{the}\mathrm{locus}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{of}$$$$\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{lines}AN\mathrm{and}BM.$$

Answered by ajfour last updated on 16/Dec/17

Answered by ajfour last updated on 16/Dec/17

$$letM(a,0)andN(0,b)$$$$slopeofMN=-\frac{1}{slopeofAB}$$$$\Rightarrow -\frac{\mathit{b}}{\mathit{a}}=\frac{3}{4}\dots \left(\mathit{i}\right)$$$$slopeofAP=slopeofAN$$$$\Rightarrow \frac{k-0}{h-(-3)}=\frac{b}{3}$$$$\Rightarrow \frac{\mathit{k}}{\mathit{h}+3}=\frac{\mathit{b}}{3}\dots .\left(\mathit{i}\mathit{i}\right)$$$$slopeofBP=slopeofBM$$$$\Rightarrow \frac{k-(-4)}{h-0}=\frac{4}{a}$$$$\Rightarrow \frac{\mathit{k}+4}{\mathit{h}}=\frac{4}{\mathit{a}}\dots .\left(\mathit{i}\mathit{i}\mathit{i}\right)$$$$\left(ii\right)\times \left(iii\right)gives:$$$$\frac{k(k+4)}{h(h+3)}=\frac{4b}{3a}$$$$forlocusofP,h\to xandk\to y$$$$\Rightarrow \frac{y(y+4)}{x(x+3)}=-1\left[using\left(i\right)\right]$$$$or{\mathit{x}}^2+{\mathit{y}}^2+3\mathit{x}+4\mathit{y}=0.$$

Commented by Tinkutara last updated on 16/Dec/17

Thank you Sir!

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