Login
Login

Question

Shortest distance between the parabola y2 = 4x and x2 + y2 – 4x – 16y + 64 = 0 is equal to

Options :

  1. 2√3 – 2

  2. 3√2 – 3

  3. 4√5 – 2

  4. 2√5 – 2

Show Answer

Answer :

2√5 – 2

Solution :

N : y = mx – 2am – am3

N : y = mx – 2m – m3

It passes through (2, 8)

8 = 2m – 2m – m3

⇒ m = – 2

∴  N : y + 2x = 12

Point of intersection of normal with y2 = 4x is (4, 4)

∴ Shortest distance =

( 4 − 2 ) 2 + ( 4 − 8 ) 2 − 4 + 64 − 64

= 20 − 2

= 2 5 − 2

Report
More Similar Tests

Related Tests

Card image
  • JEE
  • intermediate
  • 3 hours
  • physics, chemistry, mathematics

JEE Advanced Paper 2 Mock Test

Prepare for success with our JEE Advanced Paper 2 Mock Test, designed to simulate the actual exam experience. This test covers Physics, Chemistry, and Mathematics with a balanced mix of conceptual and application-based questions. Sharpen your problem-solving skills and time management with carefully curated questions that align with the latest JEE syllabus. Boost your confidence and gauge your readiness for the big day!

Sophia Jaymes
By Physics Philosopher
Card image
  • JEE
  • intermediate
  • 3 hours
  • physics, chemistry, mathematics

JEE Advanced Paper 1 Mock Test

Prepare for the JEE Advanced with our comprehensive Paper 1 Mock Test, designed to simulate the actual exam experience. Sharpen your problem-solving skills with a challenging mix of Physics, Chemistry, and Mathematics questions tailored to test your knowledge and speed.

Sophia Jaymes
By Physics Philosopher

Copyright © 2025 Test Academy All Rights Reserved