A) \[{{x}^{2}}+{{y}^{2}}-5x+7y=0\]
B) \[{{x}^{2}}+{{y}^{2}}+5x-7y=0\]
C) \[{{x}^{2}}+{{y}^{2}}+5x+7y=0\]
D) \[{{x}^{2}}+{{y}^{2}}-5x-7y=0\]
Correct Answer: D
Solution :
Let the equation of PQ be \[\frac{x}{a}+\frac{y}{b}=1\] ...(i) | ||
Then equation of BP is \[\frac{x}{a}+\frac{y}{7}=1\] | ...(ii) | |
and equation of AQ is \[\frac{x}{5}+\frac{y}{b}=1\] | ...(iii) | |
Since \[AB\bot PQ,\] therefore, \[-\frac{b}{a}\times -\frac{7}{5}=-1\] \[\Rightarrow \]\[5a+7b=0\] | ...(iv) | |
Now, slope of \[BP\,\operatorname{is}-\frac{7}{a}\]and slope of \[AQ\]is \[-\frac{b}{5}\] |
So, product of these slopes is \[\frac{7b}{5a}=-1\] [from (iv)] |
\[\therefore BP\,\operatorname{and}\,AQ\].intersect each other at right angle. So, the locus of the point of intersection of \[BP\]and \[AQ\] is a circle with \[AB\] as diameter. Thus equation of locus is |
\[x\left( x-5 \right)+y\left( y-7 \right)=0\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}-5x-7y=0\] |
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