CAT 2019 Set-2 | Question: 87

Question

Let $a,b,x,y$ be real numbers such that $a^{2}+b^{2}=25,x^{2}+y^{2}=169$, and $ax+by=65$. If $k=ay-bx$, then

• A. $k=0$
• B. $0< k\leq \frac{5}{13}$
• C. $k=\frac{5}{13}$
• D. $k> \frac{5}{13}$

Answer 1

Given that,

• $a^{2}+b^{2}=25 \quad \longrightarrow (1)$
• $x^{2}+y^{2}=169 \quad \longrightarrow (2) $
• $ax+by = 65 \quad \longrightarrow (3)$
• $ k=ay-bx \quad \longrightarrow (4)$

Multiply equation $(1)$ and equation $(2)$, we get

$(a^{2}+b^{2}) (x^{2}+y^{2}) = 25 \ast 169$

$\Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = 5^{2} \ast 13^{2}$

$ \Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2} y^{2}=(5 \ast 13)^{2}$

$ \Rightarrow a^{2}x^{2} +a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = (65)^{2}$

$ \Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = (ax+by)^{2} \quad \left[\because \text {From equation (3)} \right]$

$ \Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = a^{2}x^{2} + b^{2}y^{2} + 2abxy $

$ \Rightarrow a^{2}y^{2} + b^{2}x^{2} – 2abxy = 0$

$ \Rightarrow (ay – bx)^{2} = 0$

$ \Rightarrow ay – bx = 0 $

$ \Rightarrow \boxed{k=0} \quad \left[\because \text {From equation (4)}\right]$

Correct Answer : A