GATE 2024 EE GA Q.5: solution and answer

GATE 2024 · EEGA Q.5 · 1 mark · MCQ

If, for non-zero real variables $x, y$ , and real parameter $a > 1$ ,

x : y = (a + 1) : (a - 1),

then, the ratio $(x^{2} - y^{2}) : (x^{2} + y^{2})$ is

(A)
$2 a : (a^{2} + 1)$
(B)
$a : (a^{2} + 1)$
(C)
$2 a : (a^{2} - 1)$
(D)
$a : (a^{2} - 1)$

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Answer and solution

ANSWER

$(x^{2} - y^{2}) : (x^{2} + y^{2}) = 2 a : (a^{2} + 1)$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

Write the ratio with a common multiplier: $x = k (a + 1)$ , $y = k (a - 1)$ for some non-zero real $k$ . (Because $a > 1$ , $a - 1 \neq = 0$ , so both $x$ and $y$ are non-zero, as the question says.)
Then $x + y = 2 ak$ and $x - y = 2 k$ . By the difference of squares, $x^{2} - y^{2} = (x + y) (x - y)$ : $x^{2} - y^{2} = (2 ak) (2 k) = 4 a k^{2}$
Sum of squares, using $(a + 1)^{2} + (a - 1)^{2} = (a^{2} + 2 a + 1) + (a^{2} - 2 a + 1) = 2 a^{2} + 2$ : $x^{2} + y^{2} = k^{2} (2 a^{2} + 2) = 2 k^{2} (a^{2} + 1)$
Divide. $k^{2}$ cancels, and the denominator is never zero: $\frac{x ^{2} - y ^{2}}{x ^{2} + y ^{2}} = \frac{4 a k ^{2}}{2 k ^{2} ( a ^{2} + 1 )} = \frac{2 a}{a ^{2} + 1}$
So $(x^{2} - y^{2}) : (x^{2} + y^{2}) = 2 a : (a^{2} + 1)$ , which is option (A).
Check with $a = 2$ : $x : y = 3 : 1$ , so $(9 - 1) : (9 + 1) = 8 : 10 = 4 : 5$ , and $2 a : (a^{2} + 1) = 4 : 5$ . They match.

KEY IDEA

A ratio fixes $x$ and $y$ only up to a common factor. Write $x = k (a + 1)$ and $y = k (a - 1)$ . Both $x^{2} - y^{2}$ and $x^{2} + y^{2}$ are homogeneous of degree 2, so $k^{2}$ cancels. You can also use componendo-dividendo: from $x^{2} : y^{2} = (a + 1)^{2} : (a - 1)^{2}$ you get $(x^{2} - y^{2}) : (x^{2} + y^{2}) = 4 a : (2 a^{2} + 2) = 2 a : (a^{2} + 1)$ .

WATCH OUT FOR

Forgetting the cross terms: $(a + 1)^{2} + (a - 1)^{2}$ is $2 a^{2} + 2$ , because the $+ 2 a$ and $- 2 a$ cancel. It is not $2 a^{2}$ and it is not $(2 a)^{2}$ . Another slip is dropping the factor 2 in $x - y = 2 k$ , which leads to option (B), $a : (a^{2} + 1)$ . Mixing up $x^{2} - y^{2}$ with $x^{2} + y^{2}$ (or $x + y$ with $x - y$ ) leads to the $a^{2} - 1$ options, (C) and (D).

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GATE 2024 EE GA Q.5: solution and answer

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