Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The focal chord to    is tangent to <img alt="(x-6)^2+y^2=2" src="https://entranc

The focal chord to  y^2=16 x  is tangent to (x-6)^2+y^2=2   then the possible values of the slope of this chord are

Option: 1

\begin{aligned} & (-1,1) \\ \end{aligned}


Option: 2

(-2,2) \\


Option: 3

\left(-2, \frac{1}{2}\right) \\


Option: 4

\left(2,-\frac{1}{2}\right)


Answers (1)

best_answer

Here, the focal chord of y^2=16 x is tangent to circle (x-6)^2+y^2=2.

⇒ Focus of parabola as (a, 0) \text { i.e. }(4,0) \text {. }.

Now, tangents are drawn from (4,0) \text { to }(x-6)^2+y^2=2

Since, PA is tangent to the circle.

\tan \theta = slope of tangent    =\frac{A C}{A P}=\frac{\sqrt{2}}{\sqrt{2}}=1 \quad \frac{B C}{B P}=-1

∴ Slope of focal chord as tangent to circle  = \pm 1=(-1,1)

Posted by

Ritika Jonwal

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question