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

In this article, we have solved 6 Long answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.

$1(i)~$ Find the equation of the parabola whose vertex is $~(2, 3)~$ and the equation of latus rectum is $~x=4.~$ Find the co-ordinates of the point of intersection of this parabola with its latus rectum.

Solution.

The equation of the straight line passing through the point $~(2,3)~$ and perpendicular to $~x=4~$ is $~y=4.$

So, the co-ordinates of the focus of the parabola are $~(4,3).$

$\therefore~ 2+a=4 \Rightarrow a=2.$

So, the required equation of parabola is $~(y-3)^2=8(x-2).$

Now, the co-ordinates of the point of intersection of the parabola with its latus rectum are given by $~(2+2, 3+2\times 2)=(4,7)~$ and $~(2+2,3-2\times 2)=(4,-1).$

$(ii)~$ The co-ordinates of the two ends of latus rectum of a parabola are $~(8, 1)~$ and $~(-4, 1);~$ find the equation of the parabola.

Solution.

Since the ordinates of the extremities of the latus rectum are equal, so the latus rectum is parallel to $~x-$ axis and the axis of the parabola is parallel to the $~y-$ axis.

So, the equation of latus rectum is $~y=1,~$ and co-ordinates of focus are given by the mid-point of $~(-4,1)~$ and $~(8,1)~$ which are $~\left(\frac{8-4}{2},\frac{1+1}{2}\right)=(2,1).$

Let the axis of the parabola be $~x=k(\neq 0)~$ which passes through $~(2,1).~$ So, the axis of the parabola is $~x=2.$

$~\text{Now,}~4a\\=~\text{length of the latus rectum}\\=\sqrt{(8+4)^2+(1-1)^2}\\=12 \\ \therefore~ a=\frac{12}{4}=3.$

The co-ordinates of the vertex of parabola can be $~(2,1+a)=(2,1+3)=(2,4)~$ and another possible co-ordinates of the vertex of the parabola can be $~(2,1-a)=(2,1-3)=(2,-2).$

So, the equation of the parabola with vertex $~(2,-2)~$ is

$(x-2)^2=4 \times 3(y+2) \\ \text{or,}~~ x^2-4x+4=12y+24 \\ \text{or,}~~ x^2-4x-12y=20.$

Again, the equation of the parabola with vertex $~(2,4)~$ is

$~ (x-2)^2=-12(y-4) \\ \text{or,}~~ x^2-4x+4=-12y+48 \\ \text{or,}~~ x^2-4x+12y=44.$

$2.~$ Find the equation of the parabola whose vertex is $~(-2, 2)~$ and focus is $~(-6, -6).$

Solution.

The equation of the axis of the parabola is the straight line joining the points $~(-6,-6)~$ and $~(-2,2)~$ and is given by

$~\frac{y-2}{-6-2}=\frac{x+2}{-6+2} \\ \text{or,}~~ \frac{y-2}{-8}=\frac{x+2}{-4} \\ \text{or,}~~ \frac 12(y-2)=x+2 \\ \text{or,}~~ y-2=2(x+2) \\ \text{or,}~~ 2x-y+6=0 \rightarrow(1)$

The equation of the directrix which is perpendicular to the straight line $~(1)~$ is $~~x+2y=k \rightarrow(2)$

Now, the co-ordinates of the focus $~(S)\equiv (-6,-6),~$ the co-ordinates of the vertex $~(A)\equiv (-2,2).$

Let the point of intersection of the axis of the parabola and the directrix be $~N(p,q).$

Since $~A~$ is the mid-point of $~S~$ and $~N~$ , so

$~\frac{-6+p}{2}=-2 \Rightarrow p=2,~~\frac{-6+q}{2}=2 \Rightarrow q=10.$

$\therefore~ N \equiv (2,10).$

Since the straight line $~(2)~$ passes through $~(2,10),~$

$~2 + 2\times 10=k \Rightarrow k=22.$

Hence by $~(2),~$ we get the equation of the directrix $~ x+2y=22.$

Let $~P(x,y)~$ be any point on the parabola.

$~PM \\= \text{length of the perpendicular from P upon the directrix}\\=\frac{|x+2y-22|}{\sqrt{1^2+2^2}}\\=\frac{|x+2y-22|}{\sqrt{5}}.$

Since $~P~$ lies on the required parabola, hence we must have

$~\overline{SP}=\overline{PM} \\ \text{or,}~~ SP^2=PM^2 \\ \text{or,}~~ (x+6)^2+(y+6)^2=\frac 15 (x+2y-22)^2 \\ \text{or,}~~ 5[(x+6)^2+(y+6)^2]=(x+2y-22)^2\rightarrow(3)$

Hence, $~(3)~$ represents the required equation of parabola.

$3~.$ The directrix of a parabola is $~x+y+4=0~$ and vertex is the point $~(-1, -1).~$ Find $~(i)~$ the position of focus and $~(ii)~$ the equation of the parabola. [Council Sample Question ‘ $13$ ]

Solution.

The equation of the directrix of the parabola is $~x+y+4=0 \rightarrow(1).$

The axis of the parabola is perpendicular to the directrix and is passing through the vertex $~A(-1,-1).$

The equation of any straight line perpendicular to $~(1),~$ can be written as $~ x-y=k \rightarrow(2).$

Since the straight line $~(2)~$ is passing through the point $~(-1,-1)~$ ,

$~-1-(-1)=k \Rightarrow k=0.$

So, by $~(2),~$ we get the equation of the axis of the parabola which is $~x-y=0 \rightarrow(3).$

Solving $~(1)~$ and $~(3)~$ , we get $~x=-2,~~y=-2.$

So, the point of intersection of the axis of parabola and the directrix is $~N(-2,-2).$

Let $~~S(p,q)~$ be the focus of the parabola and $~A(-1,-1)~$ be the mid-point of $~N~$ and $~S.$

$\therefore~\frac{p-2}{2}=-1 \Rightarrow p=0,~~\frac{q-2}{2}=-1 \Rightarrow q=0.$

So, $~S\equiv (0,0).$

Let $~P~$ be any point on the parabola and $~PM~$ is the perpendicular distance from $~P~$ upon the directrix.

$\therefore~\overline{SP}=\overline{PM} \\ \text{or,}~~ SP^2=PM^2 \\ \text{or,}~~ x^2+y^2=\left[\frac{x+y+4}{\sqrt{1^2+1^2}}\right]^2 \\ \text{or,}~~ 2(x^2+y^2)=(x+y+4)^2 \\ \text{or,}~~ x^2+y^2-2xy-8x-8y-16=0\rightarrow(4).$

The equation $~(4)~$ represents the required parabola.

$4(i)~$ The axis of a parabola is parallel to $~x-$ axis and it passes through the points $~(2, 0), (1, -1) and (6, -2);~$ find its equation.

Solution.

Since the axis of the parabola is parallel to $~x-$ axis, the equation of the parabola can be written as $~~x=ay^2+by+c \rightarrow(1)$

Since the parabola $~(1)~$ is passing through $~(2,0),$

$~2=0+0+c \Rightarrow c=2. \\ \therefore~ x=ay^2+by+2 \rightarrow (2)$

Now, the parabola $~(2)~$ is passing through the points $~(1,-1)~$ and $~(6,-2).$

$\therefore~ 1=a(-1)^2+b(-1)+2 \\ \text{or,}~~ a-b=-1 \rightarrow(3)$

$~6=a(-2)^2+b(-2)+2 \\ \text{or,}~~ 6=4a-2b+2 \\ \text{or,}~~ 3=2a-b+1 \\ \text{or,}~~ 2a-b=2 \rightarrow(4)$

From $~(3)~$ and $~(4)~$ we get,

$~(2a-b)-(a-b)=2-(-1) \Rightarrow a=3.$

$\therefore~ 3-b=-1~~[\text{By (3)}] \\ \text{or,}~~ b=3+1=4.$

Hence, the equation of the required parabola is $~~x=3y^2+4y+2.$

$(ii)~$ A parabola passes through the points $~(0, 0), (2, 2) and (-2, -6)~$ and its axis is parallel to $y-$ axis. Find its equation.

Solution.

By the question, since the axis of the parabola is parallel to $~y-$ axis , the equation of the parabola can be written as $~y=ax^2+bx+c \rightarrow(1)$

Since the parabola $~(1)~$ passes through the point $~(0,0)~$ ,

$~0=0+0+c \Rightarrow c=0.$

So, equation $~(1),~$ can be written as $~y=ax^2+bx \rightarrow(2)$

Since the parabola $~(2)~$ passes through the points $~(2,2)~$ and $~(-2,-6),$

$~2=a\times 2^2+b \times 2 \Rightarrow 1=2a+b \rightarrow(3)$

$-6=a(-2)^2+b(-2) \\ \text{or,}~~ -6=4a-2b \\ \text{or,}~~ -3=2a-b\rightarrow(4)$

From $~(3)~$ and $~(4),~$ we get

$1-3=(2a+b)+(2a-b) \\ \text{or,}~~ -2=4a \\ \text{or,}~~ a=-\frac 24=-\frac 12.$

So, $~ 1=2(-1/2)+b \Rightarrow b=2.$

Hence, the equation of the parabola is

$~y=-\frac 12 \cdot x^2+2x \\ \text{or,}~~ 2y=-x^2+4x \\ \text{or,}~~ x^2-4x+2y=0.$

$5(i)~$ If the extremities of a focal chord of the parabola $~y^2=4ax~$ be $~(at_1^2, 2at_1)~$ and $~(at_2^2, 2at_2)~$ , prove that $~t_1t_2=-1.$

Solution.

The equation of the focal chord joining the points $~(at_1^2,2at_1)~$ and $~(at_2^2,2at_2)~$ is

$~ y-2at_1=\frac{2at_2-2at_1}{at_2^2-at_1^2} (x-at_1^2) \\ \text{or,}~~ y-at_1=\frac{2a(t_2-t_1)}{a(t_2+t_1)(t_2-t_1)} (x-at_1^2) \\ \text{or,}~~ y-2at_1=\frac{2}{t_2+t_1}(x-at_1^2) \rightarrow(1)$

Since the focal chord $~(1)~$ passes through the point $~(a,0),$

$~0-2at_1=\frac{2}{t_1+t_2}(a-at_1^2) \\ \text{or,}~~ -2at_1=\frac{2a(1-t_1^2)}{t_1+t_2} \\ \text{or,}~~ -t_1=\frac{1}{t_1+t_2}(1-t_1^2) \\ \text{or,}~~ -t_1(t_1+t_2)=1-t_1^2 \\ \text{or,}~~ -t_1^2-t_1t_2=1-t_1^2 \\ \text{or,}~~ t_1t_2=-1.$

$(ii)~$ If $~(at^2,2at)~$ be the co-ordinates of an extremity of a focal chord of the parabola $~y^2=4ax,~$ the show that the length of the chord is $~a\left(t+\frac 1t\right)^2.$

Solution.

If $(at^2,2at)~$ be the co-ordinates of an extremity of a focal chord of the parabola, another extremity is $~(a/t^2,-2a/t)~~[\because tt_1=-1]$

So, the length of the chord is

$=\sqrt{(at^2-a/t^2)^2+(2at+2a/t)^2}~~\text{unit}\\=\sqrt{a^2(t^2-1/t^2)^2+4a^2(t+1/t)^2} ~~\text{unit} \\=\sqrt{a^2(t+1/t)^2(t-1/t)^2+4a^2(t+1/t)^2} ~~\text{unit} \\=a(t+1/t)\sqrt{(t-1/t)^2+4} ~~\text{unit}\\=a(t+1/t)\sqrt{(t-1/t)^2+4 \cdot t \cdot 1/t} ~~\text{unit}\\=a(t+1/t) \sqrt{(t+1/t)^2} ~~\text{unit}\\=a(t+1/t)(t+1/t) ~~\text{unit} \\=a(t+1/t)^2 ~~\text{unit}$

$6.~$ Show that the equation of the chord of the parabola $~y^2=4ax~$ through the points $~(x_1,y_1)~$ and $~(x_2,y_2)~$ on it is $~(y-y_1)(y-y_2)=y^2-4ax.$

Solution.

Since the points $~(x_1,y_1)~$ and $~(x_2,y_2)~$ lie on the parabola $~y^2=4ax,~$ so

$~y_1^2=4ax_1 \rightarrow(1),~~y_2^2=4ax_2\rightarrow(2)$

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

$~y_1^2-y_2^2=4a(x_1-x_2) \\ \text{or,}~~ (y_1+y_2)(y_1-y_2)=4a(x_1-x_2) \\ \text{or,}~~\frac{y_1-y_2}{x_1-x_2}=\frac{4a}{y_1+y_2} \rightarrow(3)$

The equation of the straight line joining the points $~(x_1,y_1)~$ and $~(x_2,y_2)~$ is

$\frac{y-y_1}{x-x_1}=\frac{y_1-y_2}{x_1-x_2} \\ \text{or,}~~ \frac{y-y_1}{x-x_1}=\frac{4a}{y_1+y_2}~~[\text{By (3)}] \\ \text{or,}~~ (y_1+y_2)y-(y_1+y_2)y_1=4a(x-x_1) \\ \text{or,}~~ y^2-(y_1+y_2)y+y_1y_2+y_1^2=4ax_1-4ax+y^2 \\ \text{or,}~~ (y-y_1)(y-y_2)=y^2-4ax ~~[\text{By (1)}]$

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Parabola (S .N. Dey ) | Ex-4 | Part-1

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