Parabola (S .N. Dey ) | Ex-4 | Part-3

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In the previous article , we have solved 6 Long Answer Type Questions (7-12) of Parabola Chapter. In this article, we have solved 6 more Long answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.

Parabola related long answer type questions and solutions, S N Dey
13.~~ PQ~ is a double ordinate of the parabola ~y^2 = 4ax;~ find the equation to the locus of its point of trisection.

Solution.

parabola, s n dey math solution, class 11

Since ~PQ~ is a double ordinate of the parabola ~y^2=4ax,~ let ~P \equiv(at^2,2at)~,~Q \equiv (at^2,-2at).

Let the chord ~PQ~ is trisected at the points ~M~ and ~M'~, where ~M \equiv (at^2,2at/3)~,~M'\equiv (at^2,-2at/3).

\text{Let}~(h,k) =(at^2,2at/3) \\ \therefore~~ h=at^2 \rightarrow(1),\\~ k=2at/3 \Rightarrow t=\frac{3k}{2a} \rightarrow(2).

So, by ~(1),~(2)~ we get,

~h=a(3k/2a)^2 =a \times \frac{9k^2}{4a^2}=\frac{9k^2}{4a} \\ \text{or,}~~ 9k^2=4ah.

Hence, the equation of the required locus of the point of trisection is ~~9y^2=4ax.

14.~ Show that the circle described on a focal chord of a parabola as diameter touches its directrix.

Solution.

Let the equation of the parabola be ~y^2=4ax \rightarrow(1). 

We know the extremities of the focal chord of the parabola ~(1)~ be ~(at^2,2at)~ and ~\left(a/t^2,-2a/t\right)~.

\therefore~ the equation of the circle having diameter as the focal chord with the aforesaid extremities is 

~(x-at^2)(x-a/t^2)+(y-2at)(y+2a/t)=0 \\ \text{or,}~~ x^2+y^2+x(-a/t^2-at^2)+y(2a/t-2at)-3a^2=0\rightarrow(2)

Now, the equation of the directrix of the parabola is ~x+a=0 \rightarrow(3)

From ~(2)~ and ~(3)~ we get,

~a^2+y^2+a(a/t^2+at^2)+y(2a/t-2at)-3a^2=0\\ \text{or,}~~ y^2-2y(at-a/t)+(a^2t^2-2 \cdot at \cdot a/t+ a^2/t^2)=0 \\ \text{or,}~~ y^2-2y(at-a/t)+(at-a/t)^2=0 \\ \text{or,}~~ y=(at-a/t)

So, the point of intersection of the straight line ~(3)~ and the circle ~(2)~ is ~\left(-a,at-\frac at\right).

Since ~y~ has only one value, so we can say that the circle touches its directrix.

15.~ Prove that the sum of the reciprocals of the segments of any focal chord of a parabola is constant.

Solution.

Let the equation of the parabola be ~y^2=4ax~ and ~PP'~ be the focal chord of the parabola.

~l=\text{distance between the focus}~(a,0)~\text{and}~P(at^2,2at)\\=\sqrt{(at^2-a)^2+(2at-0)^2}\\=\sqrt{a^2(t^2-1)^2+4a^2t^2}\\=a\sqrt{(t^2-1)^2+4t^2}\\=a\sqrt{(t^2+1)^2}\\=a(t^2+1)

~l'=\text{distance between the focus}~(a,0)~\text{and}~P'(at_1^2,2at_1)\\=a(t_1^2+1)\\=a\left(\frac{1}{t^2}+1\right)~~[\because~ tt_1=-1]\\=a \left(\frac{t^2+1}{t^2}\right)\\~~\\~\therefore~~\frac 1l+\frac{1}{l'}\\=\frac{1}{a(t^2+1)}+\frac{t^2}{a(t^2+1)}\\=\frac{1+t^2}{a(t^2+1)}\\=\frac 1a=\text{constant}.

16.~ The length of latus rectum of a parabola is ~16~ unit. The distance of a point ~P~ on the parabola from its axis is ~12~ unit. Find the distance of ~P~ from the focus of the parabola.

Solution.

Let the equation of the parabola : ~y^2=4ax.

The length of the latus rectum , ~4a=16 \Rightarrow a=\frac{16}{4}=4.

Any point on the parabola can be written as ~(at^2,2at).

By question, ~2at=12 \Rightarrow t=\frac{12}{2 \times 4}=\frac 32.

\therefore~~(at^2,2at)=\left(4 \times \frac 94, 2 \times 4 \times \frac 32\right)=(9,12).

\therefore~ The distance between the focus ~(4,0)~ and ~(9,12)

=\sqrt{(9-4)^2+(12-0)^2}=\sqrt{25+144}=\sqrt{169}=13~~\text{unit.}

17.~ Prove that the length of any chord of the parabola ~y^2=4ax~ passing through the vertex and making an angle ~\theta~ with the positive direction of the ~x- axis is ~4a\csc\theta\cot\theta.

Solution.

Let ~OP~ be the chord passing through the vertex ~O,~ where ~P \equiv (at^2,2at).

By question, the slope of ~OP : \tan\theta =\frac{2at}{at^2}=\frac 2t \\ \text{or,}~~ t=2\cot\theta.

\therefore~P \equiv (4a\cot^2\theta,4a\cot\theta).

\therefore~ the length of the chord ~OP

=\sqrt{16a^2\cot^4\theta+16a^2\cot^2\theta}\\=4a\cot\theta\sqrt{\cot^2\theta+1}\\=4a\cot\theta\csc\theta.

Here, ~\csc\theta =\text{cosec}~\theta.

18.~ If ~a \neq 0~ and the line ~2bx+2cy+4d=0~ is passing through the points of intersection of parabolas ~y^2 = 4ax~ and ~x^2 = 4ay,~ then prove that ~d^2 = a^2(2b+3c)^2.

Solution.

The points of intersection of the parabolas ~y^2=4ax~ and ~x^2=4ay~ are ~(0,0)~ and ~(4a,4a).

Now, if the straight line ~2bx+3cy+4d=0~ passes through  the point ~(0,0)~, then ~d=0~  which is impossible as ~d \neq 0.

So, the straight line ~2bx+3cy+4d=0~ passes through  the point ~(4a,4a)~.

\therefore~~ 2b \times 4a+3c \times 4a+4d=0 \\ \text{or,}~~ 8ab+12ac+4d=0 \\ \text{or,}~~ 4(2ab+3ac+d)=0 \\ \text{or,}~~d=-a(2b+3c) \\ \therefore~~ d^2=a^2(2b+3c)^2~~\text{(proved)}

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