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QJEEM24Maacc8: JEE Main 2024 Mathematics - Class 11 JEEM24Maacc8 Solution: A variable line passes through the point and intersects the…

A variable line L\mathrm{L} passes through the point (3,5)(3,5) and intersects the positive coordinate axes at the points A\mathrm{A} and B\mathrm{B} . The minimum area of the triangle OAB\mathrm{OAB} , where O\mathrm{O} is the origin, is :

Multiple Choice Options

  • A 35
  • B 25
  • C 30
  • D 40

Step-by-Step Verified Solution

JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 20 English Explanation

amp;xa+yb=1amp;3a+5b=1amp;3b+5a=abamp;5a−ab=−3bamp;a(5−b)=−3bamp;a=3bb−5\begin{aligned}& \frac{x}{a}+\frac{y}{b}=1 \\& \frac{3}{a}+\frac{5}{b}=1 \\& 3 b+5 a=a b \\& 5 a-a b=-3 b \\& a(5-b)=-3 b \\& a=\frac{3 b}{b-5}\end{aligned}

amp; Area of triangle =∣12×a×b∣amp;=12×3bb−5×bamp;⇒f(b)=3b22b−10amp;⇒f′(b)=0,(2b−10)6b−2(3b2)=0amp;12b2−60b−6b2=0amp;6b2−60b=0amp;b2−10b=0amp;b(b−10)=0amp;b=0 or b=10\begin{aligned}& \text { Area of triangle }=\left|\frac{1}{2} \times a \times b\right| \\& =\frac{1}{2} \times \frac{3 b}{b-5} \times b \\& \Rightarrow f(b)=\frac{3 b^2}{2 b-10} \\& \Rightarrow f^{\prime}(b)=0,(2 b-10) 6 b-2\left(3 b^2\right)=0 \\& 12 b^2-60 b-6 b^2=0 \\& 6 b^2-60 b=0 \\& b^2-10 b=0 \\& b(b-10)=0 \\& b=0 \text { or } b=10\end{aligned}

So for minimum area, b=10b=10

then 12×a×b=30\frac{1}{2} \times a \times b=30

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