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JEE Main 202124 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2021 · 24 February · Evening Shift · Mathematics · Application of Derivatives

If the area of the triangle formed by the x -axis, the normal and...

If the area of the triangle formed by the xx-axis, the normal and the tangent to the circle (x−2)2+(y−3)2=25{\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 25 at the point \left(5, 7\right)isisA, then 24A24A is equal to _________.If the area of the triangle formed by the x -axis, the normal and the tangent to the circle (x - 2) ^ 2 + (y - 3) ^ 2 =

Correct answer

0

Step-by-step solution

If the area of the triangle formed by the x -axis, the normal and the tangent to the circle (x - 2) ^ 2 + (y - 3) ^ 2 =

Equation of normal

4x−3y+1=04x - 3y + 1 = 0

and equation of tangents

3x+4y−43=03x + 4y - 43 = 0

Area of triangle=12(433+14)×(7)= \dfrac{1}{2}\left(\dfrac{43}{3} + \dfrac{1}{4}\right) \times \left(7\right)

=12(172+312)×7= \dfrac{1}{2}\left(\dfrac{172 + 3}{12}\right) \times 7

A=122524A = \dfrac{1225}{24} Sq. unit

24A=122524A = 1225

Note: As positive xx-axis was given in the question so question was bonus in JEE Mains.

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