isi-entrance 2011 Q3

isi-entrance · India · solved Conic sections Chord Properties and Midpoint Problems

Let a line with slope of $60 ^ { \circ }$ be drawn through the focus $F$ of the parabola $y ^ { 2 } = 8 ( x + 2 )$. If the two points of intersection of the line with the parabola are $A$ and $B$ and the perpendicular bisector of the chord $A B$ intersects the $x$-axis at the point $P$, then the length of the segment PF is
(a) $16 / 3$
(b) $8 / 3$
(c) $16 \sqrt{3} / 3$
(d) $8 \sqrt{3}$

(A) The centre of the parabola is at $( - 2,0 )$. Comparing the equation of parabola with $y ^ { 2 } = 4 a ( x - a )$ we get, $a = 2$. Hence, coordinate of focus $F$ is $( 0,0 )$. The equation of the line passing through focus and having slope $\tan 60 ^ { \circ }$ is, $y = \sqrt{3} x$. Solving with parabola: $x = 4 , - 4 / 3$. Hence coordinate of $A$ and $B$ are $( 4,4 \sqrt{3} )$ and $( - 4 / 3 , - 4 \sqrt{3} / 3 )$. Midpoint of $AB = ( 4 / 3,4 \sqrt{3} / 3 )$, slope of perpendicular bisector $= - 1 / \sqrt{3}$. Putting $y = 0$ gives $x = 16/3$. Distance between $F(0,0)$ and $P(16/3,0)$ is $16/3$.

Let a line with slope of $60 ^ { \circ }$ be drawn through the focus $F$ of the parabola $y ^ { 2 } = 8 ( x + 2 )$. If the two points of intersection of the line with the parabola are $A$ and $B$ and the perpendicular bisector of the chord $A B$ intersects the $x$-axis at the point $P$, then the length of the segment PF is\\
(a) $16 / 3$\\
(b) $8 / 3$\\
(c) $16 \sqrt{3} / 3$\\
(d) $8 \sqrt{3}$

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