Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the equation of parabola whose latus rectum is 5 units, the axis is the line <img alt="6 x+8 y-4=0" src="https://entrancecorner.oncodecogs.com/gif.latex?6%20x&plus;8%20y-4%3

Find the equation of parabola whose latus rectum is 5 units, the axis is the line 6 x+8 y-4=0 and tangent at the vertex is the line 8 x-6 y+7=0.

Option: 1

(6 x+8 y-4)^2=50(8 x+6 y+7)


Option: 2

(6 x+8 y-4)^2=5(8 x-6 y+7)


Option: 3

(6 x+8 y-4)^2=50(8 x-6 y+7)


Option: 4

None of these


Answers (1)

best_answer

Equation of parabola when equation of axis, tangent at vertex and latusrectum are given

Let the equation of axis be lx + my + n = 0 and tangent at the vertex be mx - ly + r = 0.

 

Equation of parabola is

\\\mathrm{\;\;\;\;\;\;\;\;\;(PM)^2=(Latusrectum)\cdot (PN)}\\\\\mathrm{\Rightarrow \left ( \frac{lx+my+n}{\sqrt{l^2+m^2}} \right )^2=(Latusrectum)\cdot\left ( \frac{mx-ly+r}{\sqrt{m^2+n^2}} \right )}

 

Using the above result, the equation of required parabola is

\\ \left(\frac{|6 h+8 k-4 |}{\sqrt{6^{2}+8^{2}}}\right)^{2}=5\cdot\frac{8 h-6 k+7}{\sqrt{6^{2}+8^2}}\\\\(6 h+8 k-4)^2=50(8 h-6 k+7)

 

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question