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.

$2(i)~$ Find the equation of the parabola whose vertex is at $~(2,-3)~$ , axis is parallel to $~x-$ axis and length of the latus rectum is $~12.$

Solution.

We know that the equation of the parabola with the vertex $~(\alpha,\beta)~$ can be written as $~(y-\beta)^2= \pm 4a(x-\alpha),~$ where $~4a~$ is the length of the latus rectum and the axis of the parabola is parallel to $~x-$ axis.

Here, $~(\alpha,\beta)=(2,-3),~~4a=12~\text{unit}.$

Hence, the required equation of the parabola is

$~(y+3)^2= \pm 12(x-2).$

$(ii)~$ The co-ordinates of the vertex and focus of a parabola are $~(1,2)~$ and $~(-1,2)~$ respectively; find its equation.

Solution.

$~a=\text{Distance between the vertex and focus}\\~~=\sqrt{(1+1)^2+(2-2)^2}=2$

Here, Vertex $\rightarrow(1,2)~$ . Clearly the focus and vertex of the parabola lie on $~y=2~$ which is parallel to $~x-$ axis.

Since the focus lies on the left side of the vertex, the equation of the parabola is $~(y-\beta)^2=-4a(x-\alpha).$

Here, $~(\alpha,\beta) =(1,2),~~a=2.$

So, the required equation of the parabola is

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

$(iii)~$ Show that the equation of the parabola whose vertex is $~(2,3)~$ and focus is $~(2,-1)~$ is $~x^2-4x+16y=44.$

Solution.

$~a=\text{Distance between the vertex and focus}\\~~~=\sqrt{(2-2)^2+(3+1)^2}=4$

Here, Vertex $(\alpha,\beta)\rightarrow(2,3)~$ . Clearly the focus and vertex of the parabola lie on $~x=2~$ which is parallel to $~y-$ axis.

Since the focus lies below the vertex, the equation of the parabola is $~(x-\alpha)^2=-4a(y-\beta).$

Here, $~(\alpha,\beta) =(2,3),~~a=4.$

So, the required equation of the parabola is

$~(x-2)^2=-4 \times 4(y-3) \\ \text{or,}~~ (x-2)^2+16(y-3)=0 \\ \text{or,}~~ x^2-4x+16y=44.$

$3.~$ Show that the equation of the parabola whose vertex and focus are on the $~x-$ axis at distances $~a~$ and $~a'~$ from the origin respectively is $~y^2=4(a'-a)(x-a).$

Solution.

Clearly, the vertex of the parabola $~(a,0).~$ Again, the vertex and focus of the parabola lie on the $~x-$ axis . So, the axis of the parabola lies along the $~x-$ axis.

Now, the equation of the parabola can be written as

$~(y-\beta)^2=\pm 4A(x-\alpha)\rightarrow(1),~~$ where $~4A~$ is the length of the latus rectum.

By question, $~(\alpha,\beta)=(a,0)\rightarrow(2).$

So, by $~(1)~$ and $~(2),~$ we can write the equation of the parabola as $~~y^2=\pm 4A(x-a)$

Now, $~4A=4 \text{(Distance between the vertex and focus)}\\~~=4|a-a'|.$

$\therefore~ y^2= \pm 4|a-a'|(x-a)$

Case-1 :

Now, if $~a'<a~$ , focus lies on the left side of the vertex of the parabola.

$\therefore~$ the equation of the parabola

$~y^2=-4|a-a'|(x-a) \\ \text{or,}~~y^2=-4(a-a')(x-a)~\\~~[\because ~a'<a \Rightarrow |a-a'|=a-a'.] \\ \therefore~~y^2=4(a'-a)(x-a).$

Case-2 :

Again, for $~a'>a~$ , focus lies on the right side of the vertex of the parabola.

$~y^2=4|a-a'|(x-a) \\ \text{or,}~~y^2=-4(a-a')(x-a)\\~~[\because ~a'>a \Rightarrow |a-a'|=-(a-a').] \\ \therefore~~y^2=4(a'-a)(x-a).$

Hence, for both cases, the equation of the parabola is $~y^2=4(a'-a)(x-a).$

$4(i)~$ Find the equation of the parabola whose vertex is the point $~(-2,3)~$ and directrix is the line $~x+7=0.$

Solution.

The directrix of the parabola is the line $~x+7=0~$ which is parallel to the $~y-$ axis. So, the axis of the parabola is parallel to $~x-$ axis.

Again, since the vertex of the parabola is on the right side of the directrix, the equation of the parabola can be written as

$~(y-\beta)^2=4a(x-\alpha) \\ \text{or,}~~ (y-3)^2=4a(x+2)\\~~[~~\because~~\text{Vertex} (\alpha,\beta)=(-2,3)]$

Now, $~a=\text{The distance between vertex and the directrix} \\ \text{or,}~~ a=\frac{|-2+7|}{\sqrt{1^2+0^2}}=5.$

Hence, the equation of the required parabola is

$(y-3)^2=4 \times 5(x+2) \\ \text{or,}~~(y-3)^2=20(x+2).$

$(ii)~$ Find the equation of the parabola whose vertex is the point $~(1,-2)~$ and the equation of directrix is $~y+5=0$

Solution.

The equation of the directrix is $~y=-5.~$ The co-ordinates of the vertex is $~(1,-2).$

$\text{Now,}~a\\=\text{the distance between the vertex and directrix}\\=-2-(-5)\\=3.$

So, the equation of the parabola is

$(x-1)^2=4 \times 3 \times [y-(-2)] \\ \text{or,}~~ (x-1)^2=12(y+2).$

Parabola (S .N. Dey ) | Ex-4 | Part-5

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