Question #123536
Let y^2=4ax be a parabola and p be a point on it. Let the normal at p intersects the x-axis at Q. Draw a line at Q perpendicular to the above normal. Show that the line intersects the parabola y^2+4a(x-2a)=0
1
Expert's answer
2020-06-22T18:19:54-0400

let us consider the point P(a,2a) on the parabola y2=4ax.y^2=4ax.

The rough diagram of the question is attached here...




from P the normal to the parabola y2=4axy^2=4ax is drawn which intersects the x axis at the point Q,the coordinate of which is (3a,0). ( working of Q is shown below).

from Q a line perpendicular to PQ is drawn and we have to show that it intersects the parabola

y2+4a(x2a)=0.y^2+4a(x-2a)=0.


Now equation of tangent at the point P(a,2a) is given by

y.2a=2a(x+a)or y=x+a  ..............(1)y.2a=2a(x+a)\\or \space y=x+a \space \space ..............(1) (Here slope (m) of tangent is 1.)

\therefore slope of the Normal is 1.-1.

Now Equation of Normal (which is perpendicular to the tangent at P ) is given by

(y2a)=1(xa)or y=x+3a.(y-2a)=-1(x-a)\\or \space y=-x+3a.\\

AS this normal intersects X-axis at Q. so ordinate of Q must be '0'.

putting y=0 in y=x+3a.y=-x+3a.

we get x=3a.

\therefore The coordinate of Q is (3a,0).


Now slope of the line which is passing through Q (3a,0) and perpendicular to the normal at P

is again 11 as product of slope of two perpendiculars line is always 1-1 .


\therefore The equation of line passing through Q(3a,0) is given by

(y0)=1(x3a)or y=x3a.(y-0)=1(x-3a)\\or \space y=x-3a.

Now the coordinates of the end points of the latus rectum of the parabola y2=4ax isy^2=4ax \space is \\

P(a,2a) and(a,2a).P(a,2a) \space and (a,-2a).


clearly the Coordinate (a,-2a) satisfies both y=x3a and y2+4a(x2a)=0.y=x-3a \space and \space y^2+4a(x-2a)=0.

Hence the required line (y=x3a)(y=x-3a) intersects the parabola y2+4a(x2a)=0.y^2+4a(x-2a)=0.

(PROVED).

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