Its parabola.

Geometry Level 3

Consider a parabola $(x-7)^2+(y-11)^2=\dfrac{(5x-12y+6)^2}{169}$ From any point $P$ on the parabola the feet of perpendicular to the axis of the parabola is $Q$ . From the same point $P$ , the normal is drawn to the parabola to meet the axis of the parabola at point $R$ . Find length $QR$ .

The answer is 7.

3 solutions

Chaitanya Jain
Mar 12, 2014

The distance between these two points is a constant which is equal to 2a and it is called the subnormal .

a = distance of the focus from the vertex = distance of the vertex from the directrix of the parabola.

Therefore,

length of the subnormal = 2a = distance of the focus from the directrix.

By looking at the equation, it can be concluded that the focus is (7,11) and the directrix is: 5x-12y+6=0

So the distance of the focus from the directrix is

{5(7)-12(11)+6} / {5^2 +12^2}^1/2

= 91/13

QR=2a=7

Good Solution , I did it the same way

Harsh Depal - 7 years, 3 months ago

Gurjeet Singh
Mar 16, 2014

distance from point (7, 11) to the line(Directrix) 5x-12y+6=0 and than use property of parabola. QR=ZS=7

Vinay Kumar Agarwal
Mar 10, 2014

IT IS GIVEN A FIXED LINE AND A FIXED POINT i.e. FOCUS . THE DISTANCE BETWEEN THESE TWO IS EQUAL TO 2a =7. NOW CONSIDER A STANDARD PARABOLA HAVING 2a =7 units y^{2}= 2 x 2ax ; < y^{2}>=14x ; NOW CONSIDER A GENERAL POINT ON THE PARABOLA (at^{2},2at) EQUATION OF NORMAL ON THIS POINT IS GIVEN BY y= -tx + 2at +at^{3} ; BUT ITS Y CO ORDINATE IS ZERO AT THE AXIS 0= -tx + 2at +at^{3} ; x= 7 + (7/2)t^{2} ; NOW EQUATION OF PERPENDICULAR ON THE AXIS IS GIVEN BY x= (7/2)t^{2} ; SO THE DISTANCE BETWEEN THESE TWO POINTS IS ; 7 + (7/2)t^{2} - (7/2)t^{2} ; =7 SO THE ANSWER IS EQUAL TO 7

Its parabola.

The answer is 7.

3 solutions

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