Parabola (S .N. Dey ) | Ex-4 | Part-7

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In the previous article, we have solved few Short answer type questions (5-10) of Parabola Chapter . In this article, we have solved Short answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.
Parabola (S .N. Dey ) -Complete Solution

11.~ Show that the locus of the middle points of chords of the parabola ~x^2=4ay~ passing through the vertex is the parabola ~x^2=2ay.

Solution.

One end of the chord is the vertex of parabola, which is  ~O(0,0).

Let the mid point of the chords be ~M(h,k)~ and the other end of the chord be ~P(2at,at^2).

Mid-point of ~OP~ is ~\left(\frac{2at+0}{2},\frac{at^2+0}{2}\right)

\therefore~M \equiv \left(at,\frac 12at^2\right)=(h,k)

\text{So,}~h=at\rightarrow(1),~~\\ k=\frac 12 at^2=\frac 12 a(h/a)^2~~[\text{By (1)}] \\ \text{or,}~~ k=\frac{h^2}{2a} \Rightarrow h^2=2ka \rightarrow(2).

12.~ Find the locus of the middle points of a family of focal chords of the parabola ~y^2=4ax.

Solution.

The extremities of the focal chord of the parabola ~y^2=4ax~ are ~(at^2,2at)~ and ~\left(\frac{a}{t^2},-\frac{2a}{t}\right).

Let the mid-point of the chord be ~(h,k).

\therefore~h=\frac{at^2+\frac{a}{t^2}}{2} \\ \text{or,}~~ t^2+\frac{1}{t^2}=\frac{2h}{a} \\ \text{or,}~~ \left(t-\frac 1t\right)^2+2 \cdot t \cdot \frac 1t=\frac{2h}{a} \\ \text{or,}~~ \left(t-\frac 1t\right)^2=\frac{2h}{a}-2 \rightarrow(1) \\ ~~k=\frac{2at-\frac{2a}{t}}{2}=at-\frac at \\ \text{or,}~~ \frac ka=t-\frac 1t \rightarrow(2)

From ~(1)~ and ~(2)~ we get,

~\left(\frac ka\right)^2=\frac{2h}{a}-2 \\ \text{or,}~~ \frac{k^2}{a^2}=\frac{2(h-a)}{a} \\ \text{or,}~~ k^2=2a(h-a).

Hence, the locus of the mid-point ~(h,k)~ of the chord is 

y^2=2a(x-a) \\ \text{or,}~~ y^2-2ax+2a^2=0. 

13.~~PN~ is any ordinate of the parabola ~y^2=4ax;~ the point ~M~ divides ~PN~ in the ratio ~m:n.~ Find the locus of ~M.

Solution.

Let ~P(x_1,y_1)~ be any point on the parabola and ~PN~ be the ordinate of the parabola.

\therefore~N\equiv (x_1,0)~~\text{and}~~y_1^2=4ax_1 \rightarrow(1)

Again, let ~M(h,k)~ divides ~PN~ in the ratio ~m:n.

\therefore~~h=\frac{mx_1+nx_1}{m+n},~~k=\frac{m \times 0+ny_1}{m+n} \\ \text{or,}~~ h=x_1,~~ k=\frac{ny_1}{m+n}\Rightarrow ~y_1=\frac{k(m+n)}{n}

\text{Again,}~~y_1^2=4ax_1 \\ \text{or,}~~ \frac{k^2(m+n)^2}{n^2}=4ah \\ \text{or,}~~ (m+n)^2k^2=4an^2h

\therefore~ the locus of the point ~M(h,k)~ is ~(m+n)^2y^2=4an^2x.

14.~ Prove that the lines joining the ends of latus rectum of the parabola ~y^2=4ax~ with the point of intersection of its axis and directrix are at right angles.

Solution.

Let ~S(a,0)~ be the focus and ~AB~ be the length of the latus rectum of length ~4a~ unit of the parabola ~y^2=4ax.

\therefore~ A \equiv (a,2a),~B\equiv (a,-2a).

The directrix of the parabola ~x=-a~ and the axis of the parabola ~y=0.

So, the point of intersection of the directrix and the axis of the parabola is ~C(-a,0).

Slope ~(m_1)~ of ~AC~ is ~\frac{2a-0}{a+a}=1.

Slope ~(m_2)~ of ~BC~ is ~\frac{-2a-0}{a+a}=-1.

So, ~m_1 \times m_2=1 \times (-1)=-1.

Hence, ~AC \perp ~BC.

15.~ Prove that the lines joining the ends of latus rectum of the parabola ~y^2=4ax. The ordinate of ~P~ is twice that of ~Q.~ Prove that the locus of the mid-point of ~PQ~ is ~5y^2=18ax.  

Solution.

Let ~Q \equiv(x_1,y_1),~~P \equiv(x_2,y_2).

By question, ~y_2=2y_1 \rightarrow(1)

Now, since ~P,~Q~ lies on the parabola , 

~y_1^2=4ax_1 \rightarrow(2),~~y_2^2=4ax_2 \rightarrow(3).

Let ~R(h,k)~ be the mid-point of ~PQ.

~\therefore~~h=\frac{x_1+x_2}{2} \Rightarrow x_1+x_2=2h\rightarrow(4)

~\text{Again,}~~k=\frac{y_1+y_2}{2}=\frac{3y_1}{2}~~[\text{By (1)}] \\ \text{or,}~~ y_1=\frac{2k}{3}\rightarrow(5)

By ~(2)+(3)~, we get

~y_1^2+y_2^2=4a(x_1+x_2) \\ \text{or,}~~ y_1^2+4y_1^2=4a(x_1+x_2)\\ \text{or,}~~ 5(2k/3)^2=4a \times 2h~~[\text{By (4),~(5)}] \\ \text{or,}~~ 5k^2=18ah.

Hence, the locus of ~R(h,k)~ is ~5y^2=18ax.

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