Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let P be the point of intersection of the common tangents to the parabola and the

Let P be the point of intersection of the common tangents to the parabola \mathrm{y^2=12 x} and the hyperbola \mathrm{8 x^2-y^2=8}. If S and \mathrm{S^{\prime}}  denote the foci of the parabola and hyperbola respectively where \mathrm{S^{\prime}} lies on the negative x-axis, then P divides \mathrm{SS^{\prime}}in a ratio
 

Option: 1

2: 1
 


Option: 2

5: 4
 


Option: 3

14: 13


Option: 4


13: 11


Answers (1)

best_answer

Equation of any tangent to \mathrm{y^2=12 x } is

\mathrm{y=m x+\frac{3}{m} \quad \quad \quad (i) }
Also, equation of any tangent to \mathrm{ \frac{x^2}{1}-\frac{y^2}{8}=1 }is

\mathrm{ y=m x \pm \sqrt{m^2-8} \quad \quad \quad (ii) }
Since (i) and (ii) are common tangents

\mathrm{ \therefore \frac{3}{m}= \pm \sqrt{m^2-8} }

\mathrm{ \Rightarrow m^4-8 m^2-9=0 \Rightarrow\left(m^2-9\right)\left(m^2+1\right)=0}
\mathrm{ \Rightarrow m^2=9 \Rightarrow m= \pm 3 \quad [Rejecting m^2=-1 ]}
\mathrm{ \therefore}  Equations of tangents are\mathrm{ y=3 x+1\quad \quad \quad (iii) }
and \mathrm{ y=3 x-1 \quad \quad \quad (iv) }
Solving (iii) and (iv), we get a point of intersection, i.c.,

\mathrm{ P\left(-\frac{1}{3}, 0\right) }
Also, we have, \mathrm{ \frac{x^2}{1}-\frac{y^2}{8}=1}
Here, \mathrm{ c^2=\left(1+\frac{b^2}{a^2}\right)=1+8=9 \Rightarrow c= \pm 3}
So, the coordinates of foci are \mathrm{ ( \pm a e, 0)=( \pm 3,0)}



\mathrm{ ( \therefore \text { Required ratio }=\frac{P S}{P S^{\prime}}=\sqrt{\frac{\left(3+\frac{1}{3}\right)^2}{\left(-\frac{1}{3}+3\right)^2}}=\frac{\frac{10}{\frac{3}{8}}}{\frac{3}{3}}=\frac{5}{4} }

Posted by

mansi

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

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question