Question

The locus of the midpoint of the portion between the axes of
$xcos\alpha +ysin\alpha =p,$ where p is a constant, is

• A. $x^2+y^2=\frac{4}{p^2}$
• B. $x^2+y^2=4p^2$
• C. $\frac{1}{x^2}\frac{1}{y^2}=\frac{2}{p^2}$
• D. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$

Answer 1

The correct option is D $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$

$TheequationofABisxcos\alpha +ysin\alpha =p\Rightarrow \frac{xcos\alpha }{p}+\frac{ysin\alpha }{p}=1\Rightarrow \frac{x}{p/cos\alpha }+\frac{y}{p/sin\alpha }=1$
So the cordinates of $AandBare(p/cos\alpha ,0)and(0,p/sin\alpha )$
therefore, the coordinates of the midpoint of $AB$ are $(\frac{p}{2cos\alpha },\frac{p}{2sin\alpha })=(x_1,y_1)$



$x_1=\frac{p}{2cos\alpha }and{y}_1=\frac{p}{2sin\alpha }\therefore cos\alpha =p/2x_{1}andsin\alpha =p/2y_{1}co{s}^2\alpha +si{n}^2\alpha =1\Rightarrow \frac{p^2}{4}(\frac{1}{x_1^2}+\frac{1}{y_1^2})=1thelocusof(x_1,y_1)is\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$