Hyperbola (S.N.Dey) |Part-2

In the previous article , we have solved few VSA type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few more questions of Hyperbola related problems.

Hyperbola-VSA type Questions — Hyperbola-VSA Type Questions’ Solution

9(i) If the length of latus rectum of a rectangular hyperbola is $~6~$ unit, find its equation.

Solution.

The length of the latus rectum is $~\frac{2b^2}{a}=6 \rightarrow(1).$

For rectangular hyperbola, $~a=b.$

So, from $(1)$ we get,

$\frac{2a^2}{a}=6 \Rightarrow a=3.$

Hence, the equation of the rectangular hyperbola is given by

$x^2-y^2=a^2 \Rightarrow x^2-y^2=3^2=9.$

(ii) Find the co-ordinates of the foci of the rectangular hyperbola $~x^2-y^2=9.$

Solution.

Comparing the given hyperbola $~x^2-y^2=3^2~$ with $~x^2-y^2=a^2~$ we get, $~a=3.~$ Also, we know that the eccentricity $(e)$ of any rectangular hyperbola is $~\sqrt{2}.$

The co-ordinates of the foci is given by

$(\pm ae,0)=(\pm 3 \sqrt{2},0).$

10. If the latus rectum and the transverse axis of a hyperbola are equal, show that it is a rectangular hyperbola.

Solution.

We know that the length of latus rectum of the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\rightarrow(1)~$ is $~\frac{2b^2}{a}~$ unit and the transverse axis is given by $~2a~$ unit.

By question,

$\frac{2b^2}{a}=2a \Rightarrow b^2=a^2.$

Hence, by $(1)$ we get,

$\frac{x^2}{a^2}-\frac{y^2}{a^2}=1 \Rightarrow x^2-y^2=a^2.$

So, the given hyperbola is a rectangular hyperbola.

11. Find the positions of the points with respect to the hyperbola $~2x^2-y^2=7.$

$(i)~~(3,-2)~~(ii)~(4,5)~~(iii)~(-2,3)$

Solution(i)

We know that the point $~(x_1,y_1)~$ lies outside, on or inside the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ according as

$~\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}-1<, = ~~\text{or,}~~>0 \rightarrow(1).$

The given equation of the hyperbola can be written as

$~\frac{x^2}{7/2}-\frac{y^2}{7}=1.$

Now, for the point $~(3,-2)~$ we get,

$\frac{3^2}{7/2}-\frac{(-2)^2}{7}-1=\frac{18}{7}-\frac 47-1=\frac{18-4-7}{7}=1>0\rightarrow(2)$

Hence, by $(1)$ and $(2)$ we can conclude that the point $(3,-2)$ lies inside the given hyperbola.

Solution(ii)

We know that the point $~(x_1,y_1)~$ lies outside, on or inside the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ according as

$~\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}-1<, = ~~\text{or,}~~>0 \rightarrow(1).$

The given equation of the hyperbola can be written as

$~\frac{x^2}{7/2}-\frac{y^2}{7}=1.$

Now, for the point $~(4,5)~$ we get,

$\frac{4^2}{7/2}-\frac{5^2}{7}-1=\frac{32}{7}-\frac {25}{7}-1=\frac{32-25}{7}-1=1-1=0\rightarrow(2)$

Hence, by $(1)$ and $(2)$ we can conclude that the point $(4,5)$ lies inside the given hyperbola.

Solution(iii)

We know that the point $~(x_1,y_1)~$ lies outside, on or inside the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ according as

$~\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}-1<, = ~~\text{or,}~~>0 \rightarrow(1).$

The given equation of the hyperbola can be written as

$~\frac{x^2}{7/2}-\frac{y^2}{7}=1.$

Now, for the point $~(-2,3)~$ we get,

$\frac{(-2)^2}{7/2}-\frac{3^2}{7}-1=\frac{8}{7}-\frac {9}{7}-1=\frac{8-9}{7}-1=-\frac 87<0\rightarrow(2)$

Hence, by $(1)$ and $(2)$ we can conclude that the point $(-2,3)$ lies outside the given hyperbola.

12. Find the position of the point $~(7,2)~$ with respect to the hyperbola $~9x^2-16y^2=144.$

Solution.

We know that the point $~(x_1,y_1)~$ lies outside, on or inside the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ according as

$~\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}-1<, = ~~\text{or,}~~>0 \rightarrow(1).$

The given equation of the hyperbola can be written as

$~\frac{x^2}{16}-\frac{y^2}{9}=1.$

Now, for the point $~(7,2)~$ we get,

$\frac{7^2}{16}-\frac{2^2}{9}-1=\frac{49}{16}-\frac 49-1=\frac{233}{144}>0\rightarrow(2)$

Hence, by $(1)$ and $(2)$ we can conclude that the point $(7,2)$ lies inside the given hyperbola.

13. Show that the locus of the point of intersection of the lines $~\frac xa-\frac yb=m~$ and $~m\left(\frac xa+\frac yb\right)=1,~m$ being a variable parameter, is a hyperbola.

Solution.

The given equations of straight lines are

$~\frac xa-\frac yb=m\rightarrow(1),~m\left(\frac xa+\frac yb\right)=1\rightarrow(2)$

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

$~\frac xa-\frac yb=m \\ \text{or,}~~\left(\frac xa+\frac yb\right)\left(\frac xa-\frac yb\right)=m\left(\frac xa+\frac yb\right) \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\rightarrow(3)$

Hence, by $(3)$ we can conclude that the locus of point of intersection of the given lines is a hyperbola.

14. Find the parametric co-ordinates of the point $~\left(\frac{1}{\sqrt{3}},\frac 12\right)~$ on the hyperbola $~12x^2-4y^2=3.$

Solution.

The equation of the hyperbola can be written as

$\frac{12x^2}{3}-\frac{4y^2}{3}=1 \\ \text{or,}~~ 4x^2-\frac{4y^2}{3}=1 \\ \text{or,}~~\frac{x^2}{(1/2)^2}-\frac{y^2}{(\sqrt{3}/2)^2}=1 \rightarrow(1)$

Comparing $(1)$ with $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ we get,

$~a=\frac 12,~~b=\frac{\sqrt{3}}{2}.$

So, we can say that the any point on the hyperbola $(1)$ can be written in parametric form as $~(a\sec\theta,b\tan\theta) \equiv\left(\frac 12 \sec\theta,\frac{\sqrt{3}}{2}\tan\theta\right).$

$\therefore~$ By question,

$\frac 12 \sec\theta=\frac{1}{\sqrt{3}} \\ \text{or,}~~ 2\cos\theta=\sqrt{3} \\ \text{or,}~~ \cos\theta=\frac{\sqrt{3}}{2}=\cos 30^{\circ} \\ \therefore~ \theta=30^{\circ}.$

Hence, the parametric co-ordinates of the given point can be written as

$~\left(\frac 12 \sec 30^{\circ},\frac{\sqrt{3}}{2}\tan 30^{\circ}\right).$

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Hyperbola (S.N.Dey) |Part-2 | Ex-6

In the previous article , we have solved few VSA type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few more questions of Hyperbola related problems.

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