Mathematics · Hyperbola

JEE Main 2024 — 4 April, Shift 1 — Question 25

Let A be a square matrix of order 2 such that $|A|=2$ and the sum of its diagonal elements is -3 . If the points $(x, y)$ satisfying $A^{2}+x A+y I=0$ lie on a hyperbola, whose transverse axis is parallel to the x -axis, eccentricity is e and the length of the latus rectum is $\ell$ , then $\mathrm{e}^{4}+\ell^{4}$ is equal to $\qquad$

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

\text{Given:\; } A\; \text{ is\; a\; } 2\times 2 \; \text{ matrix\; with \; } |A|=2, \ \operatorname{tr}(A)=-3

\text{Characteristic\; equation:\; } \lambda^2 - (\operatorname{tr}A)\lambda + |A|=0

\Rightarrow \lambda^2 + 3\lambda + 2 = 0

\text{By\; Cayley–Hamilton\; theorem:\; } A^2 + 3A + 2I = 0

\text{Given\; equation:\; } A^2 + xA + yI = 0

\text{Subtract\; (1):\; } (x-3)A + (y-2)I = 0

\text{Taking\; determinant:\; } \det\big((x-3)A + (y-2)I\big) = 0

\text{Let\; eigenvalues\; of\; } A \text{ be\; } \lambda_1, \lambda_2

\Rightarrow \big((x-3)\lambda_1 + (y-2)\big)\big((x-3)\lambda_2 + (y-2)\big) = 0

\text{Using\; } \lambda_1+\lambda_2=-3, \ \lambda_1\lambda_2=2

\Rightarrow (y-2)^2 + 3(x-3)(y-2) + 2(x-3)^2 = 0

\text{Let\; } X = x-3, \ Y = y-2

\Rightarrow Y^2 + 3XY + 2X^2 = 0

\Rightarrow (Y+X)(Y+2X) = 0

\text{This\; represents\; a\; hyperbola\; with\; transverse\; axis\; parallel\; to\; the\; } x\text{-axis.}

\text{For\; the\; hyperbola:\; } a^2 = 1, \ b^2 = 1

\text{Eccentricity:\; } e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{2}

\text{Latus\; rectum:\; } \ell = \frac{2b^2}{a} = 2

\Rightarrow e^4 + \ell^4 = (\sqrt{2})^4 + 2^4 = 4 + 16 = 20

\boxed{20}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola