WBJEE · Maths · Parabola
Let the tangent and normal at any point \(P\left(at^2\right.\), 2at), \((a > 0)\), on the parabola \(y^2=4 a x\) meet the axis of the parabola at \(T\) and \(\mathrm{G}\) respectively. Then the radius of the circle through \(\mathrm{P}, \mathrm{T}\) and \(\mathrm{G}\) is
- A \(a\left(1+t^2\right)\)
- B \(\left(1+t^2\right)\)
- C \(a\left(1-t^2\right)\)
- D \(\left(1-\mathrm{t}^2\right)\)
Answer & Solution
Correct Answer
(A) \(a\left(1+t^2\right)\)
Step-by-step Solution
Detailed explanation
\(P\left(a t^2, 2 a t\right), T\left(-a t^2, 0\right), G\left(2 a+a t^2, 0\right)\) Slope of \(P T \times\) Slope of \(P G=-1\) \(\therefore\) TG is a diameter of the circle through points \(\mathrm{P}, \mathrm{T}\) and \(\mathrm{G}\).…
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