Hyperbola (S.N.Dey) | Part-4

In the previous article , we have solved few Short Answer type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few more Short Answer Type questions of Hyperbola related problems of s n dey mathematics class 11.

Hyperbola- S N Dey Mathematics — **Hyperbola (S.N.Dey) | Part-4 | Ex-6**

4(i) For what value of $m$ the hyperbola $3x^2-my^2=48$ will pass through the point $~(8,-6)~?$ Find its eccentricity and the length of latus rectum.

Solution.

Since the hyperbola $~3x^2-my^2=48~$ passes through the point $~(8,-6)~$ ,

$3 \times 8^2-m \times (-6)^2=48 \\ \text{or,}~~ 192-36m=48 \\ \text{or,}~~ 36m=192-48 \\ \therefore~ m=\frac{144}{36}=4.$

So, the equation of the given hyperbola is

$~3x^2-4y^2=48 \\ \text{or,}~~ \frac{x^2}{16}-\frac{y^2}{12}=1 \rightarrow(1)$

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

$~a^2=16 \Rightarrow a=4,~b^2=12 \Rightarrow b=\sqrt{12}=2\sqrt{3}.$

$\therefore~$ The eccentricity $(e)$ of the hyperbola is

$~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{12}{16}}=\sqrt{\frac{28}{16}}=\sqrt{\frac 74}=\frac{\sqrt{7}}{2}.$

The length of the latus rectum is

$\frac{2b^2}{a}=\frac{2\times 12}{4}=6~~\text{unit}$

(ii) The hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point of intersection of the lines $~x+y=3,~2x-3y=1~$ and its eccentricity is $~\sqrt{3}$ ; show that its length of latus rectum is $~2\sqrt{14}.$

Solution.

$x+y-3=0 \rightarrow(1),~2x-3y-1=0 \rightarrow(2)$

So, from $(1)$ and $(2)$ we get,

$~2(x+y-3)-(2x-3y-1)=0 \\ \text{or,}~~ 2y-6+3y+1=0 \\ \text{or,}~~ 5y=6-1 \\ \text{or,}~~ y=\frac 55=1$

$\therefore~$ By $(1)$ we get,

$~x+1-3=0 \Rightarrow x=3-1=2$

So, the point of intersection of $(1)$ and $(2)$ is $~(2,1).$

Since the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point $~(2,1),~$

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

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

By $~(1) \times (2)~$ we get,

$~1=2(4-a^2) \\ \text{or,}~~ 1=8-2a^2 \\ \text{or,}~~ 2a^2=8-1 \\ \text{or,}~~ a=\sqrt{\frac 72}$

$\therefore~ b^2=2a^2=2 \times \frac 72=7$

Hence, the length of its latus rectum is

$\frac{2b^2}{a}=\frac{2 \times 7}{\sqrt{\frac 72}}=2 \times 7 \times \sqrt{\frac 27}=2\sqrt{14}.$

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(iii) The hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point of intersection of the lines $~7x+13y-87=0~$ and $~5x-8y+7=0~$ and its latus rectum is $~\frac{32\sqrt{2}}{5}.$ Find $a$ and $b$ .

Solution.

$7x+13y-87=0 \rightarrow(1),~~5x-8y+7=0 \rightarrow(2)$

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

$~5(7x+13y-87)-7(5x-8y+7)=0 \\ \text{or,}~~ 65y-435+56y-49=0 \\ \text{or,}~~ 121y=435+49 \\ \text{or,}~~ y=\frac{484}{121}=4$

From $(1)$ we get,

$~7x+13 \times 4-87=0 \\ \text{or,}~~ 7x+52-87=0 \\ \text{or,}~~7x=35 \\ \therefore~x=\frac{35}{7}=5$

So, the point of intersection of the straight lines $(1)$ and $(2)$ is $(5,4).$

Since the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point $~(5,4),~$

$~\frac{5^2}{a^2}-\frac{4^2}{b^2}=1 \\ \text{or,}~~ \frac{25}{a^2}-\frac{16}{b^2}=1 \\ \text{or,}~~ 25-\frac{16a^2}{b^2}=a^2 \\ \text{or,}~~ \frac{16a^2}{b^2}=25-a^2\rightarrow(3)$

The length of latus rectum $~\frac{2b^2}{a}=\frac{32\sqrt{2}}{5}\rightarrow(4)$

By $~(3) \times (4)~$ we get,

$\frac{16a^2}{b^2} \times \frac{2b^2}{a}=(25-a^2) \times \frac{32\sqrt{2}}{5} \\ \text{or,}~~ 5a=\sqrt{2}(25-a^2) \\ \text{or,}~~ \sqrt{2}a^2+5a-25\sqrt{2}=0 \\ \text{or,}~~a=\frac{-5 \pm \sqrt{5^2-4\sqrt{2}(-25\sqrt{2})}}{2\sqrt{2}}=\frac{-5 \pm \sqrt{25+200}}{2\sqrt{2}} \\ \therefore a=\frac{-5+\sqrt{225}}{2\sqrt{2}}~~( \because~ a>0) \\ \text{or,}~~ a=\frac{-5+15}{2\sqrt{2}}=\frac{10}{2\sqrt{2}}=\frac{5}{\sqrt{2}}=\frac{5\sqrt{2}}{2}$

By $(2)$ we get,

$~b^2=\frac{16\sqrt{2}}{5} \times \frac{5\sqrt{2}}{2}=16 \\ \therefore~ b=\sqrt{16}=4$

(iv) The hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point $~(-3,2)~$ and its eccentricity is $~\sqrt{\frac 53};$ find the length of its latus rectum.

Solution.

Since the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point $~(-3,2),~$

$\therefore~\frac{(-3)^2}{a^2}-\frac{2^2}{b^2}=1 \\ \text{or,}~~ \frac{9}{a^2}-\frac{4}{b^2}=1 \\ \text{or,}~~ 9-\frac{4a^2}{b^2}=a^2 \\ \text{or,}~~ \frac{4a^2}{b^2}=9-a^2\rightarrow(1)$

$e=\sqrt{\frac 53} \Rightarrow e^2=\frac53 \\ \text{or,}~~ 1+\frac{b^2}{a^2}=\frac 53 \\ \text{or,}~~ \frac{b^2}{a^2}=\frac 53-1=\frac 23 \rightarrow(2)$

By $~(1) \times (2)~$ we get,

$~4=\frac 23(9-a^2) \Rightarrow 12=18-2a^2 \\ \text{or,}~~ 2a^2=18-12 \\ \text{or,}~~ a^2=\frac 62=3 \\ \text{or,}~~ a=\sqrt{3}$

From $(2)$ we get, $~\frac{b^2}{3}=\frac 23 \Rightarrow b^2=2.$

$\therefore~$ the length of the latus rectum is

$\frac{2b^2}{a}=\frac{2 \times 2}{\sqrt{3}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}~~\text{unit}$

5. Find the equation of the hyperbola whose.

(i) eccentricity is $~\frac 43,~$ focus is $(0, 4)~$ and the directrix is the line $~4y-9=0.$

Solution.

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

$\therefore~$ The distance of $~P(x,y)~$ from the focus $~S(0,4)~$ is given by $~SP=\sqrt{x^2+(y-4)^2}~~\text{unit}$

Again, the distance of $~P(x,y)~$ from the directrix $~4y-9=0~$ is

$~PM=\frac{|4y-9|}{\sqrt{0^2+4^2}}=\frac{|4y-9|}{4}.$

$\because~~ e=\frac 43, \\ \therefore~~ \frac{SP}{PM}=\frac 43 \\ \text{or,}~~ \frac{SP^2}{PM^2}=\left(\frac 43\right)^2=\frac{16}{9} \\ \text{or,}~~ 9 SP^2=16 PM^2 \\ \text{or,}~~ 9[x^2+(y-4)^2]=16 \times \frac{(4y-9)^2}{16} \\ \text{or,}~~ 9(x^2+y^2-8y+16)=16y^2-72y+81 \\ \text{or,}~~ 9x^2+9y^2-16y^2=81-144 \\ \text{or,}~~ 9x^2-7y^2=-63 \\ \text{or,}~~ 7y^2-9x^2=63.$

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(ii) eccentricity is $\frac 54,~$ focus is $~(-5,0)~$ and directrix is the line $~5x+16=0.$

Solution.

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

$\therefore~$ The distance of $~P(x,y)~$ from the focus $~S(-5,0)~$ is given by $~SP=\sqrt{(x+5)^2+y^2}~~\text{unit}$

Again, the distance of $~P(x,y)~$ from the directrix $~5x+16=0~$ is

$~PM=\frac{|5x+16|}{\sqrt{5^2+0^2}}=\frac{|5x+16|}{5}.$

$\because~ e=\frac 54 , \\ \therefore ~\frac{SP}{PM}=\frac 54 \\ \text{or,}~~ \frac{SP^2}{PM^2}=\left(\frac 54\right)^2=\frac{25}{16} \\ \text{or,}~~16 SP^2=25 PM^2 \\ \text{or,}~~ 16[(x+5)^2+y^2]=25 \times \frac{(5x+16)^2}{25} \\ \text{or,}~~ 16(x^2+10x+25+y^2)=25x^2+160x+256 \\ \text{or,}~~ 16x^2+160x+400+16y^2=25x^2+160x+256 \\ \text{or,}~~16x^2-25x^2+16y^2=256-400 \\ \text{or,}~~ -9x^2+16y^2=-144 \\ \text{or,}~~ 9x^2-16y^2=144 \rightarrow(1)$

Hence, the equation $~(1)~$ represents the required equation of hyperbola.

(iii)eccentricity is $\sqrt{3},~$ focus is $~(2,3)~$ and the equation of directrix is $~x+2y=1.$

Solution.

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

$\therefore~$ The distance of $~P(x,y)~$ from the focus $~S(2,3)~$ is given by $~SP=\sqrt{(x-2)^2+(y-3)^2}~~\text{unit}$

Again, the distance of $~P(x,y)~$ from the directrix $~x+2y=1~$ is

$~PM=\frac{|x+2y-1|}{\sqrt{1^2+2^2}}=\frac{|x+2y-1|}{\sqrt{5}}.$

$\because~ e=\sqrt{3} , \\ \therefore ~~~\frac{SP}{PM}=\sqrt{3} \\ \text{or,}~~ \frac{SP^2}{PM^2}=\left(\sqrt{3}\right)^2 \\ \text{or,}~~ SP^2=3 PM^2 \\ \text{or,}~~ (x-2)^2+(y-3)^2=3 \times \frac{(x+2y-1)^2}{5} \\ \text{or,}~~5(x^2-4x+4+y^2-6y+9)\\=3[x^2+4y^2+1+2 \cdot x \cdot 2y+2 \cdot 2y \cdot(-1)+2 \cdot(-1) \cdot x] \\ \text{or,}~~ 5x^2-20x+20+5y^2-30y+45=3x^2+12y^2+3+12xy-12y-6x \\ \text{or,}~~2x^2-7y^2-12xy-14x-18y+62=0 \rightarrow(1)$

Hence, the equation $~(1)~$ represents the required equation of hyperbola.

(iv) eccentricity is $\sqrt{2},~$ focus is $~(3,1)~$ and the equation of directrix is $~2x+y-1=0.$

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

$\therefore~$ The distance of $~P(x,y)~$ from the focus $~S(3,1)~$ is given by $~SP=\sqrt{(x-3)^2+(y-1)^2}~~\text{unit}$

Again, the distance of $~P(x,y)~$ from the directrix $~2x+y-1=0~$ is

$PM=\frac{|2x+y-1|}{\sqrt{2^2+1^2}}=\frac{|2x+y-1|}{\sqrt{5}}$

$\because~ e=\sqrt{2},\\~~\therefore~~ \frac{SP}{PM}=\sqrt{2} \\ \text{or,}~~SP^2=2PM^2 \\ \text{or,}~~(x-3)^2+(y-1)^2= 2 \times \frac{(2x+y-1)^2}{5} \\ \text{or,}~~ 5(x^2-6x+9+y^2-2y+1)=2(4x^2+y^2+1+4xy-2y-4x) \\ \text{or,}~~ 5x^2-30x+45+5y^2-10y+5=8x^2+2y^2+2+8xy-4y-8x \\ \text{or,}~~ -3x^2+3y^2-22x-6y+48-8xy=0 \\ \text{or,}~~ 3x^2-3y^2+8xy+22x+6y-48=0 \rightarrow(1)$

Hence, the equation $~(1)~$ represents the required equation of hyperbola.

(v) eccentricity is $\sqrt{2},~$ focus is $~(a,0)~$ and the equation of directrix is $~x=\frac a2.$

Solution.

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

$\therefore~$ The distance of $~P(x,y)~$ from the focus $~S(a,0)~$ is given by $~SP=\sqrt{(x-a)^2+y^2}~~\text{unit}$

Again, the distance of $~P(x,y)~$ from the directrix $~x=\frac a2~$ is

$~PM=\frac{|x-a/2|}{\sqrt{1^2+0^2}}=\left|x-\frac a2\right|$

$\because~ e=\sqrt{2},\\~\therefore~~~\frac{SP}{PM}=\sqrt{2} \\ \text{or,}~~ SP^2=2PM^2 \\ \text{or,}~~ (x-a)^2+y^2=2(x-a/2)^2 \\ \text{or,}~~ x^2-2ax+a^2+y^2=2(x^2-2x \cdot a/2+a^2/4) \\ \text{or,}~~ x^2-2ax+a^2+y^2=2x^2-2ax+a^2/2 \\ \text{or,}~~ -x^2+y^2=-a^2+a^2/2 \\ \text{or,}~~-(x^2-y^2)=-\frac{a^2}{2}\\ \text{or,}~~x^2-y^2=a^2/2 \\ \therefore~2x^2-2y^2=a^2\rightarrow(1)$

Hence, the equation $~(1)~$ represents the required equation of hyperbola.

(vi) eccentricity is $~3,~$ focus is $~(-1,1)~$ and the equation of directrix is $~x-y+3=0.$

Solution.

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

$\therefore~$ The distance of $~P(x,y)~$ from the focus $~S(-1,1)~$ is given by $~SP=\sqrt{(x+1)^2+(y-1)^2}~~\text{unit}$

Again, the distance of $~P(x,y)~$ from the directrix $~x-y+3=0~$ is

$PM=\frac{|x-y+3|}{\sqrt{1^2+(-1)^2}}=\frac{|x-y+3|}{\sqrt{2}}$

$\because~ e=3,\\~~\therefore~~ \frac{SP}{PM}=3 \\ \text{or,}~~SP^2=9PM^2 \\ \text{or,}~~(x+1)^2+(y-1)^2= 9 \times \frac{(x-y+3)^2}{2} \\ \text{or,}~~ 2(x^2+2x+1+y^2-2y+1)=9(x^2+y^2+9-2xy-6y+6x) \\ \text{or,}~~ 2x^2+4x+4+2y^2-4y=9x^2+9y^2+81-18xy-54y+54x\\ \text{or,}~~ 0=7x^2+7y^2-18xy-54y+4y+54x-4x+81-4\\ \text{or,}~~ 7(x^2+y^2)-18xy-50y+50x+77=0\rightarrow(1)$

Hence, the equation $~(1)~$ represents the required equation of hyperbola.

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In the previous article , we have solved few Short Answer type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few more Short Answer Type questions of Hyperbola related problems of s n dey mathematics class 11.

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