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

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.

$1(i)~$ 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

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

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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.

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