Question

The distance between the points $(a\cos \theta +b\sin \theta ,0)and(0,a\sin \theta -b\cos \theta )$is _____.

• A. $a^2+b^2$
• B. $a+b$
• C. $a^2-b^2$
• D. $\sqrt{a^2+b^2}$

Answer 1

The correct option is D $\sqrt{a^2+b^2}$

Given the point $A(\cos \theta +b\sin \theta ,0),(0,a\sin \theta -b\cos \theta )$

By distance formula,

The distance of $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{{[0-(a\cos \theta +b\sin \theta {)}^2+(a\sin \theta -b\cos \theta )-0]}^2}=\sqrt{a^2{\cos }^2\theta +2ab\cos \theta \sin \theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta }=\sqrt{(a^2+b^2){\cos }^2\theta +(a^2+b^2){\sin }^2\theta }=\sqrt{a^2+b^2}[\because {\cos }^2\theta +{\sin }^2\theta =1]$