The perpendicular distance from (√5,tan−12) to the line 3cosθ+4sinθ+5r=0 is
A
168
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
165
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
516
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is D165 Given point (√5,tan−12) r=√5,θ=tan−12 ⇒sinθ=2√5,cosθ=1√5 Given equation is 3cosθ+4sinθ+5r=0 ⇒3rcosθ+4rsinθ+5=0 So, p=|3+8+5|5 ⇒p=165