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Maths Question 6 – JEE-MAIN 2026

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Let A=[1274−2838−7] and det(A−αI)=0, where α is a real number. If the largest possible value of α is p, then the circle (x−p)2+(y−2p)2=320, intersects the co-ordinate axes at

The values of α for which det(A−αI)=0 are the eigenvalues of matrix A.

● Video Walkthrough
Step 1: Calculate the characteristic equation and find eigenvalues✦ Active

The characteristic equation is det(A−αI)=0. For the given matrix A, we compute the determinant of A−αI:

A−αI=[1−α274−2−α838−7−α]

Expanding the determinant yields the cubic equation: α3+8α2−88α−320=0. By testing integer factors of 320, we find that α=8 is a root. Factoring the polynomial gives (α−8)(α2+16α+40)=0. The roots (eigenvalues) are α=8, α=−8+26, and α=−8−26. The largest eigenvalue is p=8 (since 26≈4.9, so −8+26≈−3.1 and −8−26≈−12.9).

Step 2: Formulate the circle equation○ Expand

Substitute p=8 into the given circle equation (x−p)2+(y−2p)2=320. This gives:

(x−8)2+(y−2×8)2=320⟹(x−8)2+(y−16)2=320

The center of the circle is (8,16) and the radius squared is r2=320.

Step 3: Find intersection points with coordinate axes○ Expand

For intersection with the x-axis, set y=0:

(x−8)2+(0−16)2=320⟹(x−8)2+256=320⟹(x−8)2=64⟹x−8=±8

This yields x=16 or x=0. So, the intersection points with the x-axis are (16,0) and (0,0). For intersection with the y-axis, set x=0:

(0−8)2+(y−16)2=320⟹64+(y−16)2=320⟹(y−16)2=256⟹y−16=±16

This yields y=32 or y=0. So, the intersection points with the y-axis are (0,32) and (0,0). The distinct intersection points with the coordinate axes are (16,0), (0,0), and (0,32). There are 3 distinct points.

💡 Teacher's Secret Hint

Remember to count only distinct points when an intersection point is common to both axes.

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