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Question Number 35696 by Tinkutara last updated on 22/May/18

Commented by Rasheed.Sindhi last updated on 22/May/18

$$\mathrm{Mr}\mathrm{Tinkutara},$$$${}^{\bullet }{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{tried}\mathrm{many}\mathrm{attempts}\mathrm{to}\mathrm{your}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{only}\mathrm{one}\mathrm{solution}\mathrm{under}\mathrm{the}\mathrm{topic},$$$${}^{\prime }\mathrm{Pattern}{\mathrm{approach}}^{\prime }.\mathrm{Other}\mathrm{attempts}$$$$\mathrm{are}\mathrm{almost}\mathrm{deleteable}.$$$${}^{\bullet }{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{produced}\mathrm{an}\mathrm{other}\mathrm{soution}\mathrm{of}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{with}\mathrm{the}\mathrm{question}.$$$$\mathrm{Pl}\mathrm{see}\mathrm{if}\mathrm{you}{\mathrm{didn}}^{\prime }\mathrm{t}.$$

Commented by ajfour last updated on 22/May/18

Commented by ajfour last updated on 22/May/18

$$2x+y=1$$$$x-2y=3$$$$intersection(h,k)$$$$h=1,k=-1$$$$\tan \theta =-2$$$$\Rightarrow \sin \theta =\frac{-2}{\sqrt{5}};\cos \theta =\frac{1}{\sqrt{5}}$$$$u=1+\frac{x-1}{\sqrt{5}}-\frac{2(y+1)}{\sqrt{5}}$$$$v=-1+\frac{2(x-1)}{\sqrt{5}}+\frac{y+1}{\sqrt{5}}$$$$\frac{{(u-h)}^2}{4}-\frac{{(v-k)}^2}{9}=1\Rightarrow 45{(u-1)}^2-20{(v+1)}^2=180$$$$9{(x-2y-3)}^2-4{(2x+y-1)}^2=180$$$$\Rightarrow \frac{{(x-2y-3)}^2}{20}-\frac{{(2x+y-1)}^2}{45}=1.$$

Commented by Tinkutara last updated on 22/May/18

@ajfour Sir: The answer is little wrong. @Rasheed Sir: I also did the first case by hit and trial only. We can do it only when c given is small.

Commented by ajfour last updated on 22/May/18

$$\mathcal{T}hanks.$$

Commented by Tinkutara last updated on 22/May/18

$$\frac{{(x-2y-3)}^2}{20}-\frac{{(2x+y-1)}^2}{45}=1$$

Commented by Tinkutara last updated on 22/May/18

Thanks Sir! It also has a shorter approach.

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