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  • Parabola and the circle having it's centre at <img alt="\mathrm{(6,5)}"

Parabola \mathrm{y^{2}=4 x} and the circle having it's centre at \mathrm{(6,5)} intersect at right angle. Possible point of intersection of these curves can be

Option: 1

(9,6)


Option: 2

(2, \sqrt{8})


Option: 3

(1,2)


Option: 4

(3,2 \sqrt{3})


Answers (1)

best_answer

Let the possible point be \mathrm{\left(\mathrm{t}^{2}, 2 \mathrm{t}\right)}

Equation of tangent at this point is
\mathrm{y t=x+t^{2}}
It must pass through \mathrm{(6,5)}
\mathrm{\Rightarrow \quad \mathrm{t}^{2}-5 \mathrm{t}+6=0}
\mathrm{\Rightarrow \quad \mathrm{t}=2,3}
Thus possible points are (4,4),(9,6).

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vinayak

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