Parabola (S .N. Dey ) | Ex-4 | Part-8

In the previous article, we have solved few Short answer type questions (11-15) of Parabola Chapter . In this article, we will solve Very Short answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.
Parabola-S.N.De Complete Solution

1(i)~ Find the focus , the length of the latus rectum and the directrix of the parabola ~3x^2=8y.

Solution.

The given parabola can be written as ~x^2=4 \cdot \frac 23\cdot y\rightarrow(1)

Comparing the  parabola ~(1)~ with ~x^2=4ay~ we get,

Focus : ~(0,a)=\left(0,\frac 23\right),~ 

Length of the latus rectum ~: (4a)=4 \times \frac 23=\frac 83~\text{unit}

The directrix of the parabola :

 ~y+a=0 \\ \text{or,}~~y+\frac 23=0 \\ \text{or,}~~ 3y+2=0.

(ii)~ Find the length of the latus rectum of the parabola ~y=-2x^2+12x-17.

Solution.

The given equation of the parabola can be written as 

~y=-2x^2+12x-17 \\ \text{or,}~~ y=-2\left(x^2-6x+\frac{17}{2}\right) \\ \text{or,}~~ y= -2 \left[x^2-2 \cdot x \cdot 3+3^2-9+\frac{17}{2}\right] \\ \text{or,}~~ y= -2 \left[(x-3)^2-\frac 12\right] \\ \text{or,}~~ y= -2(x-3)^2+1 \\ \text{or,}~~ y-1=-2(x-3)^2 \\ \text{or,}~~ (x-3)^2=-\frac 12(y-1) \\ \text{or,}~~ (x-3)^2=-4 \cdot \frac 18 (y-1)\rightarrow(1)

Hence, the length of the latus rectum of the parabola ~(1)~ is

~4a=4 \times \frac 18=\frac 12~~\text{unit.}

2(i)~ Find the equation of the parabola whose co-ordinates of the vertex and focus are ~(0,0)~ and  ~\left(\frac 32,0\right)~ respectively.

Solution.

The given parabola has the vertex ~(0,0)~ and the focus ~S(a,0) =\left(\frac 32,0\right).

Clearly, the axis of the parabola is parallel to the ~x- axis.

Hence, the equation of the parabola is 

~(y-0)^2=4 \times \frac 32(x-0) \\ \text{or,}~~ y^2=6x.

(ii)~ The vertex of a parabola is at the origin and its focus is ~\left(0,-\frac 54\right);~ find the equation of the parabola.

Solution.

The given parabola has the vertex ~(\alpha,\beta) \equiv(0,0)~ and the focus ~S(\alpha,a+\beta) =\left(0,-\frac 54\right).

Clearly, the axis of the parabola is parallel to the ~y- axis.

Hence, the equation of the parabola is 

~(x-\alpha)^2=4a(y-\beta) \\ \text{or,}~~(x-0)^2=-4 \times \frac 54(y-0) \\ \text{or,}~~ x^2=-5y \\ \text{or,}~~ x^2+5y=0.

3(i)~ The parabola ~x^2+2py=0~ passes through the point ~(4,-2);~ find the co-ordinates of focus and the length of latus rectum.

Solution.

Since the parabola ~x^2+2py=0~ passes through the point ~(4,-2),~ we have,

~4^2+2p \times (-2)=0 \\ \text{or,}~~ 4^2-4p=0 \\ \text{or,}~~ 4(4-p)=0 \\ \therefore~~ p=4.

So, the equation of the parabola turns out to be 

~x^2+2 \times 4y=0 \\ \text{or,}~~ x^2=-8y=-4 \times 2\times y \rightarrow(1)

Comparing the parabola ~(1)~ with the parabola ~x^2=-4ay~ where ~a=2~, we get the focus ~(S) : (0,-a)\equiv(0,-2)~~ and the length of the latus rectum is given by ~~4a=4 \times 2=8~~\text{unit.}

(ii)~ The parabola ~y^2=2ax~ goes through the point of intersection of ~\frac x3+\frac y2=1~ and ~\frac x2+\frac y3=1.~ Find its focus.

Solution.

~\frac x3+\frac y2=1 \rightarrow(1),~~ \frac x2+\frac y3=1 \rightarrow(2).

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

~3 \left(\frac x3+\frac y2\right)-2\left(\frac x2+\frac y3\right)=3-2 \\ \text{or,}~~ x+\frac{3y}{2}-x-\frac{2y}{3}=1 \\ \text{or,}~~ \frac{9y-4y}{6}=1 \\ \text{or,}~~ \frac{5y}{6}=1 \\ \text{or,}~~ y=\frac 65.

From ~(1)~ we get,

~\frac x3+\frac 12 \left(\frac 65\right)=1 \\ \text{or,}~~ \frac x3=1-\frac 35 =\frac 25 \\ \text{or,}~~ x=\frac 65.

Hence, the point of intersection of ~(1)~ and ~(2)~ is ~\left(\frac 65,\frac 65\right).

Since the parabola ~y^2=2ax~ passes through the point ~\left(\frac 65,\frac 65\right)~

~ \left(\frac 65\right)^2=2a\left(\frac 65 \right) \\ \text{or,}~~ 2a=\frac 65 \\ \therefore~~a=\frac 35.

So, the equation of the parabola is

~y^2=2 \times \frac 35 \times x \\ \text{or,}~~ y^2=4 \cdot \frac{3}{10} \cdot x

Hence, the focus of the parabola is  ~(a,0)=\left(\frac{3}{10},0\right).

4(i)~ A parabola having a vertex at the origin and axis along ~x-axis passes through ~(6,-2);~ find the equation of the parabola.

Solution.

The equation of the parabola having vertex at ~(0,0)~ and axis along ~x- axis can be written as 

~(y-0)^2=4a(x-0) \\ \text{or,}~~y^2=4ax \rightarrow(1)

Since the parabola ~(1)~ is passing through the point ~(6,-2)~, so we get by ~(1),

(-2)^2=4a \times 6 \Rightarrow 6a=1 \\ \therefore a=\frac 16.

Hence, the equation of the parabola ~(1)~ can be written as

~y^2=4 \times \frac 16 \times x \\ \text{or,}~~y^2=\frac 23x \\ \text{or,}~~ 3y^2=2x.

5(i)~ Find the equation of the parabola whose vertex is ~(0,0)~ and directrix is the line ~x+3=0.

Solution.

The equation of the directrix is 

x+3=0 \Rightarrow x=-3.

So, the equation of the parabola is along the positive ~x- axis and can be written as ~y^2=4ax \rightarrow(1)

Here, ~a= the distance between the directrix and the vertex of the parabola =3~ unit.

\therefore~y^2=4 \times 3 \times x=12x~~[\text{By (1)}]

(ii)~ Find the equation of the parabola whose vertex is at the origin and directrix is the line  ~y-4=0.

Solution.

The equation of the directrix is ~y-4=0 \Rightarrow y=4.

So, the equation of the directrix is along the negative ~y- axis and can be written as ~x^2=-4ay,~~ where ~a= the distance between the directrix and the vertex of the parabola =4~ unit.

\therefore~~ x^2=-4 \times 4 \times y=-16y \\ \text{or,}~~ x^2+16y=0.

6.~ If the parabola ~y^2=4ax~ passes through the point of intersection of the straight lines ~3x+y+5=0~ and ~x+3y-1=0,~ find the co-ordinates of the focus and the length of its latus rectum. 

Solution.

~~3x+y+5=0 \rightarrow(1),\\~~x+3y-1=0\rightarrow(2).

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

3x+y+5-3(x+3y-1)=0 \\ \text{or,}~~ y+5-9y+3=0 \\ \text{or,}~~-8y+8=0 \\ \text{or,}~~ 8y=8 \\ \text{or,}~~ y=\frac 88=1.

From ~(2),~~ x+3 \times 1-1=0  \\ \text{or,}~~ x+2=0 \\ \text{or,}~~ x=-2.

So, the point of intersection of the given straight lines is ~(-2,1).

Since the parabola ~y^2=4ax~ passes through the point ~(-2,1),

~1^2=4a \cdot (-2) \Rightarrow a=-\frac 18.

\therefore~~ y^2=4 \times \left(-\frac 18 \right)x

So, the co-ordinates of its focus ~~(a,0)=\left(-\frac 18,0\right).

The length of its latus rectum is 

~~4|a|=4 \times \frac 18=\frac 12~~\text{unit}.

7(i)~ The parabola ~y^2=4ax~ passes through the centre of the circle ~2x^2+2y^2-4x+12y-1=0;~ find the co-ordinates  of the focus , length of the latus rectum and the equation of the directrix.

Solution.

The equation of the given circle can be written as ~~x^2+y^2-2x+6y-\frac 12=0 \rightarrow(1)

Comparing ~(1)~ with ~x^2+y^2+2gx+2fy+x=0~, we get 

~g=-\frac 22=-1,~~f=\frac 62=3.

The centre of the given circle is ~~(-g,-f)=\left(1,-3\right).

Since the given parabola ~y^2=4ax~ passes through the point ~(1,-3)~,

~(-3)^2=4a \cdot 1 \Rightarrow a=\frac 94. 

So, the focus of the parabola ~(a,0)=\left(\frac 94,0\right).

The length of the latus rectum is 

~4a=4 \times \frac 94=9~~\text{unit.}

The equation of the directrix is given by 

~x=-a \\ \text{or,}~~ x=-\frac 94 \\ \therefore~ 4x+9=0.

(ii)~ If the parabola ~y^2=4ax~ passes through the centre of the circle ~x^2+y^2+4x-12y-4=0;~ what is the length of the latus rectum of the parabola.

Solution.

The equation of the given circle can be written as ~~x^2+y^2+4x-12y-4=0 \rightarrow(1)

Comparing ~(1)~ with ~x^2+y^2+2gx+2fy+x=0~, we get 

~2g=4 \Rightarrow g=2,~~2f=-6\Rightarrow f=-3.

The centre of the given circle is ~~(-g,-f)=\left(-2,6\right).

Since the given parabola ~y^2=4ax~ passes through the point ~(-2,6)~,

~6^2=4a \cdot (-2) \\ \text{or,}~~ a=-\frac{36}{8}=-\frac 92.

The length of the latus rectum is 

~4|a|=4 \times \frac 92=18~~\text{unit.}

8(i)~ Find the point on the parabola ~y^2=20x~ at which the ordinate is double the abscissa.

Solution.

Let ~P \equiv(at^2,2at)~ be a point on the parabola ~y^2=20x.

By question, ~2 \times at^2=2at \Rightarrow t=1.

So, ~P \equiv (at^2,2at)=(a,2a).

Since the point ~P~ lies on the given parabola,

~(2a)^2=20 \times a \Rightarrow a=\frac{20}{4}=5.

Hence, ~P \equiv(a,2a)=(5, 2 \times 5)=(5,10)~~\text{(ans.)}

(ii)~ The point on the parabola ~y^2=-36x~ at which the ordinate is three times the abscissa.

Solution.

Let ~P \equiv(at^2,2at)~ be a point on the parabola ~y^2=-36x.

By question, ~2at=3 \times at^2\Rightarrow t=\frac 23.

So, ~P \equiv (at^2,2at)=\left(a \times \frac 49,2  a\times \frac 23\right)=\left(\frac{4a}{9},\frac{4a}{3}\right).

Since the point ~P~ lies on the given parabola,

\left(\frac{4a}{9}\right)^2=-36 \times \frac{4a}{9} \\ \text{or,}~~ 4a=-36 \\ \therefore~ a=-\frac{36}{4}=-9.

\therefore~ P \equiv \left(\frac{4a}{9},\frac{4a}{3}\right)=\left(\frac 49 \times (-9),\frac 43 \times (-9)\right)=(-4,-12).

9.~ Find the point on the parabola ~y^2=4ax~(a>0)~ which forms a triangle of area ~3a^2~ with the vertex and focus of the parabola.

Solution.

Let ~P(at^2,2at)~ be a point on the parabola ~y^2=4ax~(a>0)~. Also, suppose that the vertex ~A\equiv(0,0)~ and the focus ~S \equiv (a,0).

So, area of the triangle ~\Delta~ ASP 

=\frac 12|(0+2a^2t+0)-(0+0+0)|\\=\pm \frac 12 \times 2a^2t\\=\pm a^2t

\therefore~ 3a^2=\pm a^2t \Rightarrow t=\pm 3.  

\therefore~ P \equiv (at^2,2at)=(9a, \pm 6a).

10.~ What type of conic is the locus of the moving point ~(at^2,2at)~?~ Find the equation of the locus.

Solution.

Let ~x=at^2\rightarrow(1),~~y=2at\Rightarrow t=\frac{y}{2a}\rightarrow(2)

Hence, by ~(1)~ and ~(2)~ we get,

~x=a\left(\frac{y}{2a}\right)^2 \\ \text{or,}~~ x=a \times \frac{y^2}{4a^2}=\frac{y^2}{4a} \\ \therefore~ y^2=4ax \rightarrow(3)

Hence, by ~(3)~ we can conclude that the equation of the conic represents a parabola.

11(i)~ The focal distance of a point on the parabola ~y^2=8x~ is ~4~;~ find the co-ordinates of the point.

Solution.

Let the coordinate of that point is  ~P(x,y).

Equation of the parabola ~y^2=8x =4\cdot 2 \cdot x\rightarrow(1)

Now, the co-ordinates of focus ~(S)\equiv(2,0).

By question, 

~PS=4 \\ \text{or,}~~ \sqrt{(x-2)^2+(y-0)^2}=4 \\ \text{or,}~~ (x-2)^2+y^2=4^2 \\ \text{or,}~~ x^2-4x+4+8x=16~~[\because~y^2=8x] \\ \text{or,}~~ x^2+4x-12=0 \\ \text{or,}~~ x^2+6x-2x-12=0 \\ \text{or,}~~ x(x+6)-2(x+6)=0 \\ \text{or,}~~(x+6)(x-2)=0 \\ \text{or,}~~ x=-6,2.

But since ~x~ can not be  negative, so ~x=2

So, y^2=8x=8 \times 2=16 \Rightarrow y=\pm \sqrt{16}=\pm 4.

So, the coordinates are ~(2,4),~~(2,-4).

(ii)~ Find the focal distance of a point on the parabola ~y^2=20x~ if the abscissa of the point be ~7.

Solution.

The given equation of the parabola is ~y^2=20x=4 \cdot 5 \cdot x

So, the focus ~(S)=(5,0).

If the abscissa of the point on the given parabola is ~7,~ then we need to find the ordinate of that point.

~y^2=20 \times 7=140 \Rightarrow y=\sqrt{140}=2\sqrt{35}.

Hence, the focal distance = distance between ~(5,0)~ and ~(7,2\sqrt{35})

=\sqrt{(7-5)^2+(2\sqrt{35}-0)^2}\\=\sqrt{4+140}\\=\sqrt{144}\\=12~~\text{unit.}

12(i)~ Determine the positions of the points ~(a)~(3,6)~,~ (b)~(4,3)~ and ~(c)~(1,-3)~ with respect to the parabola ~y^2=9x.

Solution.

~ C: y^2=9x ~\text{or,}~~C:y^2-9x=0

 ~C(3,6) : 6^2-9 \times 3=36-27=9>0 \rightarrow(a)

~C(4,3) : 3^2-9 \times 4=9-36=-27<0 \rightarrow(b)

~C(1,-3) : (-3)^2-9 \times 1=9-9=0 \rightarrow(c)

From ~(a),~(b),~(c)~ we can conclude that first point lies outside the parabola, second point lies inside the parabola and third point lies on the given parabola.

(ii)~ For what values of  ~a~ will the point ~(8,4)~ be an inside point of the parabola ~y^2=4ax?

Solution.

The given equation of the parabola can be written as ~y^2-4ax=0.~ We know if ~(\alpha,\beta)~ lies inside the parabola, then ~\beta^2-4a\alpha<0. 

So, if the point ~(8,4)~ lies  inside the given parabola, then

~4^2-4a \times 8<0 \\ \text{or,}~~ 16-32a<0 \\ \text{or,}~~ 16<32a \\ \text{or,}~~  32a >16 \\ \text{or,}~~  a>\frac{16}{32} =\frac 12

Leave a Comment