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

In the previous article, we have solved few Short answer type questions 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.

$5(i)~$ Find the equation of the parabola whose focus is at the origin and the equation of the directrix is $~x+y=1.$

Solution.

Focus $: S(0,0),~$ Directrix $: x+y-1=0.$

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

$\text{Now,}~~ PM\\=\text{the distance of the point P from the directrix}\\=\frac{x+y-1}{\sqrt{1^2+1^2}}\\=\frac{x+y-1}{\sqrt{2}}.$

Also, $~SP=\sqrt{x^2+y^2}.$

For parabola,

$~~~SP=PM \\ \text{or,}~~ SP^2=PM^2 \\ \text{or,}~~ x^2+y^2=\frac{(x+y-1)^2}{2} \\ \text{or,}~~ 2(x^2+y^2)=x^2+2x(y-1)+(y-1)^2 \\ \text{or,}~~ x^2+y^2-2xy+2x+2y=1.$

$(ii)~$ Find the equation of the parabola whose focus is $~(2,1)~$ and whose directrix is $~3x-y+1=0.$

Solution.

Focus : $S(2,1).~$ Directrix : $~3x-y+1=0.$

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

We know,

Distance between $~S~$ and $~P=~$ Perpendicular distance to directrix from $~P$

$\text{or,}~~\sqrt{(x-2)^2+(y-1)^2}=\frac{3x-y+1}{\sqrt{3^2+(-1)^2}} \\ \text{or,}~~ (x-2)^2+(y-1)^2 =\frac{1}{10}(3x-y+1)^2 \\ \text{or,}~~10( x^2-4x-2y+y^2+5)\\~~=9x^2+y^2+1-6xy-2y+6x \\ \text{or,}~~ 10x^2-40x-20y+10y^2+50\\~~=9x^2+y^2+1-6xy-2y+6x \\ \text{or,}~~ x^2+9y^2+6xy-46x-18y+49=0 .$

$(iii)~$ The equation of the directrix of a parabola is $~x=y~$ and the co-ordinates of its focus are $~(4,0)~$ . Find the equation of the parabola.

Focus : $S(4,0).~$ Directrix : $~x-y=0.$

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

We know,

Distance between $~S~$ and $~P=~$ Perpendicular distance to directrix from $~P$

$\text{or,}~~ \sqrt{(x-4)^2+(y-0)^2}=\frac{x-y}{\sqrt{1^2+(-1)^2}} \\ \text{or,}~~ (x-4)^2+y^2=\frac12 (x-y)^2 \\ \text{or,}~~ 2(x^2-8x+16+y^2)=x^2+y^2-2xy \\ \therefore~x^2-8x+16+y^2 x^2+y^2+2xy-16x+32=0.$

$6.~$ Find the co-ordinates of vertex and the length of latus rectum of the parabola whose focus is $~(0,0)~$ and the directrix is the line $~2x+y=1.$

Solution.

We know that the directrix $\perp~$ axis of the parabola.

Now, the slope of the directrix $=-2.$

So, the slope of the axis $=\frac 12.$

Equation of the axis which passes through the point $~(0,0)~$ and having slope $~ \frac 12~$ is $y=\frac 12 x \Rightarrow x-2y=0.$

Let $~M~$ be the point of intersection of the straight lines $~x-2y=0 \rightarrow(1)~$ and $~2x+y=1 \rightarrow(2).$

Solving $~(1)~$ and $~(2),~$ we get $~y=\frac 15,~~x=\frac 25.$

$\therefore~~ M \equiv \left(\frac 25,\frac 15\right).$

Since vertex $~(V)~$ is the mid-point of $~M~$ and $~S,~$

$~V \equiv \left(\frac 12 \times \frac 25, \frac 12 \times \frac 15\right)=\left(\frac 15,\frac{1}{10}\right).$

$~a=\text{Distance between the vertex and the focus}\\=\sqrt{(1/5-0)^2+(1/10-0)^2}=\sqrt{\frac{1}{25}+\frac{1}{100}} \\ \therefore~~ a=\sqrt{\frac{4+1}{100}}=\sqrt{\frac{5}{100}}=\frac{\sqrt{5}}{10} \\ \text{or,}~~4a=\frac{4\sqrt{5}}{10}=\frac{2}{\sqrt{5}}~~\text{unit}$

Hence, the length of the latus rectum $~(4a)=\frac{2}{\sqrt{5}}~\text{unit}.$

$7.~$ A point moves in such a way that its distance from the point $~(2,5)~$ is equal to its distance from the line $~2x+4y=3.~$ Find the equation of its path. What is the name of the curve ?

Solution.

Let at any instant the co-ordinates of the moving point $~P~$ be $~(h,k).$

The distance between the point $~(h,k)~$ and $~(2,5)~$ is

$~\sqrt{(h-2)^2+(k-5)^2}~\text{unit.}$

Again, the distance of the point $~(h,k)~$ from the from the line $~2x+4y=3~$ is

$~\frac{|2h+4k-3|}{\sqrt{2^2+4^2}}=\frac{|2h+4k-3|}{\sqrt{20}}$

By question,

$~ \sqrt{(h-2)^2+(k-5)^2}=\frac{|2h+4k-3|}{\sqrt{20}} \\ \text{or,}~~ (h-2)^2+(k-5)^2=\frac{(2h+4k-3)^2}{20} \\ \text{or,}~~ (2h+4k-3)^2=20[(h-2)^2+(k-5)^2]$

Hence, the locus of the point $~(h,k)~$ is

$~(2x+4y-3)^2=20[(x-2)^2+(y-5)^2]$ .

We know that if a point moves in such a way that it is always equidistant from a fixed point and from a fixed straight line, then the locus of the point will be parabola. So, the name of the curve is parabola.

$8.~$ The co-ordinates of a moving point $~P~$ are $~(2t^2+4,4t+6)~$ ; show that the locus of $~P~$ is a parabola.

Solution.

$\text{Clearly,}~x=2t^2+4 \rightarrow(1),\\~~y=4t+6 \Rightarrow t=\frac{y-6}{4} \rightarrow(2).$

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

$~x-4=2 \times \left(\frac{y-6}{4}\right)^2 \\ \text{or,}~~ (y-6)^2=8(x-4)~\rightarrow(3)$

Hence, by $~(3),~$ we can say that the locus of the given point is a parabola.

$9.~$ If $~\theta~$ is a variable parameter, show that the equations $~x=\frac 14(3-\csc^2\theta),~y=2+\cot\theta~$ represent the equation of a parabola. Find the co-ordinates of vertex, focus and the length of latus rectum of the parabola.

Solution.

$~x=\frac 14(3-\csc^2\theta)\\ \Rightarrow \csc^2\theta=3-4x \rightarrow(1)\\~~y=2+\cot\theta \\ \text{or,}~~ \cot\theta=y-2\rightarrow(2)$

We know that

$~1+\cot^2\theta=\csc^2\theta \\ \text{or,}~~ 1+(y-2)^2=3-4x ~~[\text{By (1),(2)}] \\ \text{or,}~~ (y-2)^2=2-4x \\ \text{or,}~~ (y-2)^2=-4\left(x-\frac 12 \right) \rightarrow(1)$

Hence, by $~(1)~$ we can conclude that the given equations represent a parabola.

Here, $~a=1.$

Vertex $\rightarrow \left(\frac 12,2\right),~$

Focus $\rightarrow \left(\frac 12-a,2\right)=\left(\frac 12-1,2\right)=\left(-\frac 12,2\right)$

Length of the latus rectum $(4a)=4 \times 1=4~~\text{unit}.$

$10.~$ Find the equation of the circle, one of whose diameters is the latus rectum of $~y^2=4ax.~$ Show that this circle goes through the common point of the axis and the directrix of the parabola.

Solution.

For the parabola $~y^2=4ax,~$ the length of the latus rectum is $~4a~$ unit and the focus is $~(a,0).$

Since the diameter of the circle is the latus rectum of the parabola, so the centre of the circle is the mid-point of the latus rectum which is $~(a,0).$

Now, the equation of the circle having centre $~(a,0)~$ and with diameter $~4a~$ is given by