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Normal at a point $P$ on the parabola $y^{2} = 4 a x$ meets
the axis at $Q$ such that the distance of $Q$ from the
focus of the parabola is 10 $a$ . The coordinates of $P$
are:

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(6a, 9a)

(6a, –9a)

(9a, 6a)

(3a, 6a)

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detailed solution

Correct option is C

Equation of the normal at Pat2,2at to the parabolay2=4ax is y=−tx+2at+at3 which meets theaxis y=0 at x=a2+t2. So coordinates of Q area2+t2,0. Focus is S(a,0)QS=2a+at2−a=a1+k2=10a⇒t2=9So the required coordinates are (9a,±6a).

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Normal at a point P on the parabola y2=4ax meetsthe axis at Q such that the distance of Q from thefocus of the parabola is 10a. The coordinates of Pare:

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Ready to Test Your Skills?

free study material

Normal at a point $P$ on the parabola $y^{2} = 4 a x$ meets
the axis at $Q$ such that the distance of $Q$ from the
focus of the parabola is 10 $a$ . The coordinates of $P$
are: