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

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In the previous article, we have solved long 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.

1.~ Find the axis, co-ordinates of vertex and focus, length of latus rectum , equation of 

directrix and the co-ordinates of the ends of latus rectum for each of the following parabolas :

(i)~ y^2=20x

Solution.

(i)~~y^2=20x=4 \times 5x \Rightarrow a=5.

So, the focus (S) \equiv (a,0)=(5,0).

Axis of the parabola \rightarrow~ positive ~x- axis.

Vertex \rightarrow (0,0).

Length of the latus rectum \rightarrow 4a=4 \times 5=20~~ \text{unit.}

The equation of the directrix ~: x+a=0 \Rightarrow x+5=0.

The extremities of latus rectum :

(5, \pm 2a)=(5, \pm 2 \times 5)=(5, \pm 10).

(ii)~ x^2=-12y

Solution.

x^2=-12y=-4 \times 3x \Rightarrow a=3.

So, the focus (S) \equiv (0,-a)=(0,-3).

Axis of the parabola \rightarrow~ negative ~x- axis.

Vertex \rightarrow (0,0).

Length of the latus rectum \rightarrow 4a=4 \times 3=12~~ \text{unit.}

The equation of the directrix ~: y-a=0 \Rightarrow y=3.

The extremities of latus rectum :

( \pm 2a,-3)=( \pm 2 \times 3, -3)=(\pm 6,-3).

(iii)~ 5x^2=16y

Solution.

5x^2=16y \\ \text{or,}~~ x^2=\frac{16}{5}y=4 \times \frac 45y. \\ \therefore~~ a=\frac 45.

So, the focus (S) \equiv (0,a)=\left(0,\frac 45\right).

Axis of the parabola \rightarrow~ positive ~y- axis.

Vertex \rightarrow (0,0).

Length of the latus rectum \rightarrow 4a=4 \times \frac 45=\frac{16}{5}~~ \text{unit.}

The equation of the directrix 

~ y=-a \\ \text{or,}~~ y=-4/5 \\ \therefore~~ 5y+4=0.

The extremities of latus rectum :

\left(\pm 2a,\frac 45\right)=\left(\pm 2 \times \frac 45, \frac 45 \right)=\left(\pm \frac{8}{5}, \frac 45\right).

(iv)~3y^2=- 4x

Solution.

3y^2=-4x \\ \text{or,}~~y^2=-\frac 43 x=-4 \times \frac 13 x \\ \therefore~a=\frac 13.

So, the focus (S) \equiv (-a,0)=\left(-\frac 13,0\right).

Axis of the parabola \rightarrow~ negative ~x- axis.

Vertex \rightarrow (0,0).

Length of the latus rectum \rightarrow 4a=4 \times \frac 13=\frac 43~~ \text{unit.}

The equation of the directrix 

~x=a \Rightarrow x=1/3 \Rightarrow 3x-1=0.

The extremities of latus rectum :

(-1/3, \pm 2a)=\left(-\frac 13, \pm \frac 23\right).

(v)~(y+3)^2=2(x+2)

Solution.

~(y+3)^2=2(x+2) \\ \text{or,}~~ (y+3)^2= 4 \times \frac 12(x+2) \\ \therefore~~ a=\frac12.

So, the focus (S) \equiv (-2+a,-3)=\left(-2+1/2,-3\right)=(-3/2,-3).

Axis of the parabola \rightarrow~ parallel to ~x- axis.

Vertex \rightarrow (-2,-3).

Length of the latus rectum \rightarrow 4a=4 \times \frac 12=2~~ \text{unit.}

The equation of the directrix 

~x=-2-a \Rightarrow x=-2-1/2 \Rightarrow 2x+5=0.

The extremities of latus rectum :

(-3/2,-3 \pm 2a)=\left(-\frac 32, -3\pm 1\right)\\ \text{i.e.,}~~(-3/2,-2),(-3/2,-4).

(vi)~4(x-2)^2=-5(y+3)

Solution.

~4(x-2)^2=-5(y+3) \\ \text{or,}~~ (x-2)^2=-\frac 54(y+3) \\ \text{or,}~~ (x-2)^2=-4 \times \frac{5}{16}(y+3). \\ \therefore~~ a=\frac{5}{16}.

So, the focus (S) \equiv

~(2,-3-a)=\left(2,-3-5/16\right)=\left(2,-\frac{53}{16}\right).

Axis of the parabola \rightarrow~ parallel to ~y- axis.

Vertex \rightarrow (2,-3).

Length of the latus rectum \rightarrow 4a=4 \times \frac{5}{16}=\frac 54~~ \text{unit.}

The equation of the directrix 

~y=-3+a \Rightarrow y=-3+\frac{5}{16} \Rightarrow 16y+43=0.

The extremities of latus rectum :

\left(2 \pm 2a,-\frac{53}{16} \right)=\left(2 \pm \frac 58, -\frac{53}{16}\right)\\ \text{i.e.,}~~\left( \frac{21}{8}, -\frac{53}{16}\right),~\left( \frac{11}{8},-\frac{53}{16}\right).

(vii)~y^2=6(x+y)

Solution.

~y^2=6(x+y) \\ \text{or,}~~ y^2-6y=6x \\ \text{or,}~~ y^2-6y+9=6x+9 \\ \text{or,}~~ (y-3)^2=6\left(x+\frac 32\right)=4 \times \frac 32(x+3/2)\rightarrow(1).

Comparing the parabola ~(1)~ with the general equation of parabola ~(y-\beta)^2=4a(x-\alpha), we get

So, the focus (S) \equiv (a+\alpha,\beta)=\left(\frac 32-\frac 32,3\right)=\left(0,3\right).

Axis of the parabola \rightarrow~ parallel to ~x- axis.

Vertex \rightarrow (-3/2,3).

Length of the latus rectum \rightarrow 4a=4 \times \frac 32=6~~ \text{unit.}

The equation of the directrix  :

x+a=\alpha \Rightarrow x+\frac 32=-\frac 32 \\ \therefore~ x+3=0.

(viii)~y^2-4x-2y-7=0

Solution.

~y^2-4x-2y-7=0 \\ \text{or,}~~ y^2-2y=4x+7 \\ \text{or,}~~ y^2-2y+1=4x+8 \\ \text{or,}~~ (y-1)^2=4 \times 1\times (x+2)\rightarrow(1) \\ \therefore~ a=1.

Comparing the parabola ~(1)~ with the general equation of parabola ~(y-\beta)^2=4a(x-\alpha), we get

So, the focus (S) \equiv

~(a+\alpha,\beta)=\left(1-2,1\right)=\left(-1,1\right).

Axis of the parabola \rightarrow~ parallel to ~x- axis.

Vertex ~(\alpha,\beta)~\rightarrow (-2,1).

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

The equation of the directrix  :

x+a=\alpha \Rightarrow x+1=-2 \\ \therefore~ x+3=0.

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