Q240 / 270 · NDA 2023 (Phase 2) · Mathematics

What is the sum of the intercepts of the line whose perpendicular distance from origin is 4 units and the angle which the normal makes with positive DIRECTIONS of $x$-axis is $15^{\circ}$ ?

  • A
    8
  • B
    $4 \sqrt{6}$
  • Correct answer
    C
    $8 \sqrt{6}$
  • D
    16

Solution

Since, equation of normal is $\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{P}$ $\Rightarrow \mathrm{x} \cos 15^{\circ}+\mathrm{y} \sin 15^{\circ}=4$...(i) If $y=0 \Rightarrow x=\frac{4}{\cos 15^{\circ}}=\mathrm{a}=$ intercept on X -axis If $x=0 \Rightarrow y=\frac{4}{\sin 15^{\circ}}=b=$ intercept on Y-axis Now, $a+b=\frac{4}{\cos 15^{\circ}}+\frac{4}{\sin 15^{\circ}}=8 \sqrt{6}$

Hence, the correct answer is option: 'C'.

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