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

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 Complete Long Answer Type questions of Hyperbola related problems of S N Dey mathematics class 11.

Hyperbola-Long Answer Type Questions — Long Answer Type Questions’ Solutions(Hyperbola)-S N Dey

1. Find the centre, the length of latus rectum, the eccentricity, the co-ordinates of foci and the equations of the directrices of the hyperbola $~\frac{(x+2)^2}{9}-\frac{(y-1)^2}{16}=1.$

Solution.

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

$~a^2=9 \Rightarrow a=3, ~b^2=16 \Rightarrow b=4,~~\alpha=-2,~~\beta=1.$

(i) The co-ordinates of centre of hyperbola : $~(\alpha,\beta)=(-2,1).$

(ii) The eccentricity $~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac 53.$

(iii) The co-ordinates of foci :

$(\alpha \pm ae,\beta)=\left(-2 \pm 3 \times \frac 53,1\right)=(-2 \pm 5,1)~$ i.e., $~(3,1)~$ and $~(-7,1).$

(iv) The equations of directrices are :

$~x= \alpha \pm \frac ae=-2 \pm \frac{3}{\frac 53}=-2 \pm \frac 95 \\ \therefore~~ 5(x+2)=\pm 9.$

2. Show that the equation $~9x^2-16y^2-18x-64y-199=0~$ represents the equation of a hyperbola ; find the co-ordinates of its centre and foci and also the equation of its directrices.

Solution.

The equation of the hyperbola can be written as

$~9(x^2-2x+1)-16(y^2+4y+4)=9-64+199 \\ \text{or,}~~9(x-1)^2-16(y+2)^2=144 \\ \text{or,}~~ \frac{(x-1)^2}{16}-\frac{(y+2)^2}{9}=1 \rightarrow(1)$

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

$~ a^2=16 \Rightarrow a=4,~~b^2=9 \Rightarrow b=3,~~\alpha=1,~~\beta=-2.$

(i) Centre : $~(\alpha,\beta)=(1,-2)$

(ii) Foci : $~(\alpha \pm ae,\beta)=\left(1 \pm 4 \times \frac 54,-2\right)=(1 \pm 5,-2)~~$ i.e., $~(6,-2)~$ and $(-4,-2).$

The eccentricity $~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac 54.$

(iii) The equations of directrices are given by

$~x=\alpha \pm \frac ae=1 \pm \frac{4}{5/4}=1 \pm \frac{16}{5} \\ \text{or,}~~ 5x=5 \pm 16 \\ \therefore~~ 5x=21,~~5x=-11\Rightarrow 5x+11=0.$

3. Find (i) the centre (ii) the vertice (iii) the equations of the axes (iv) the lengths of axes (v) the eccentricities (vi) the lengths of latera recta (vii) the co-ordinates of foci (viii) the equations of the directrices of the following two hyperbolas :

$(a)~9x^2-16y^2-90x+64y+17=0,~(b)~3x^2-3y^2-18x+12y+2=0.$

Solution.

$9x^2-16y^2-90x+64y+17=0 \\ \text{or,}~~ 9(x^2-10x+25)-16(y^2-4y+4)=225-64-17 \\ \text{or,}~~ 9(x-5)^2-16(y-2)^2=144 \\ \text{or,}~~ \frac{(x-5)^2}{16}-\frac{(y-2)^2}{9}=1 \rightarrow(1)$

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

$a^2=16 \Rightarrow a=4,~~b^2=9 \Rightarrow b=3, \alpha=5,~~\beta=2.$

(i) Centre of the hyperbola $(1)$ is $(\alpha,\beta)=(5,2)$

(ii) Co-ordinates of vertices : $(\alpha \pm a,\beta)=(5 \pm 4,2)=(9,2),~(1,2)$

(iii) The equation of transverse axis : $y-2=0~$ and conjugate axis : $~x-5=0.$

(iv)The length of transverse axis : $2a=2 \times 4=8~\text{unit}$ and the length of conjugate axis : $2b=2\times 3=6~~\text{unit}.$

(v) The eccentricity $(e)$ is given by

$~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac 54.$

(vi) The length of latera recta : $~\frac{2b^2}{a}=2 \times\frac{9}{4}=\frac 92~~\text{unit}.$

(vii) The co-ordinates of foci :

$(\alpha \pm ae,\beta)=\left(5 \pm 4 \times \frac 54,2\right)=(5 \pm 5,2).$

(viii) The equations of directrices are given by

$~x=\alpha \pm \frac ae=5 \pm \frac{4}{5/4}=5 \pm \frac{16}{5} \\ \text{or,}~~ 5x=25 \pm 16 \Rightarrow 5(x-5)=\pm 16.$

Solution (b)

$3x^2-3y^2-18x+12y+2=0 \\ \text{or,}~~3(x^2-6x+9)-3(y^2-4y+4)=-2+27-12 \\ \text{or,}~~3(x-3)^2-3(y-2)^2=13 \\ \text{or,}~~ \frac{(x-3)^2}{\frac{13}{3}}-\frac{(y-2)^2}{\frac{13}{3}}=1 \rightarrow(1)$

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

$a^2=b^2=\frac{13}{3} \Rightarrow a=b=\sqrt{\frac{13}{3}},~\alpha=3,~~\beta=2.$

(i) Centre of the hyperbola $(1)$ is $(\alpha,\beta)=(3,2)$

(ii) Co-ordinates of vertices : $(\alpha \pm a,\beta)=(3 \pm \sqrt{\frac{13}{3}},2)$

(iii) The equation of transverse axis : $y-2=0~$ and conjugate axis : $~x-3=0.$

(iv)The length of transverse axis : $2a=2 \times \sqrt{\frac{13}{3}}~\text{unit}=$ the length of conjugate axis : $2b=2\times \sqrt{\frac{13}{3}}~~\text{unit}.$

(v) The eccentricity $(e)$ is given by

$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{a^2}{a^2}}=\sqrt{2}.$

(vi) The length of latera recta : $~\frac{2b^2}{a}=2 \times\frac{\frac{13}{3}}{\sqrt{\frac{13}{3}}}=2 \cdot \sqrt{\frac{13}{3}}~~\text{unit}.$

(vii) The co-ordinates of foci :

$(\alpha \pm ae,\beta)=\left(3 \pm \sqrt{\frac{13}{3}} \times\sqrt{2},2\right)=\left(3 \pm \sqrt{\frac{26}{3}},2\right).$

(viii) The equations of directrices are given by

$~x=\alpha \pm \frac ae=3 \pm \frac{\sqrt{\frac{13}{3}}}{\sqrt{2}}=3 \pm \sqrt{\frac{13}{6}} \\ \text{or,}~~ x-3=\pm \sqrt{\frac{13}{6}}$

4. A point moves on a plane in such a manner that the difference of its distances from the points $~(4,0)~$ and $~(-4, 0)~$ is always constant and equal to $~4\sqrt{2}$ . Show that the locus of the moving point is a rectangular hyperbola whose equation you are to determine.

Solution.

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

So, by question,

$|\sqrt{(h-4)^2+k^2}-\sqrt{(h+4)^2+k^2}|=4\sqrt{2} \\ \text{or,}~~\sqrt{(h-4)^2+k^2}=\sqrt{(h+4)^2+k^2} \pm 4\sqrt{2} \\ \text{or,}~~(h-4)^2+k^2=(h+4)^2+k^2 +(4\sqrt{2})^2 \pm 8\sqrt{2} \sqrt{(h+4)^2+k^2} \\ \text{or,}~~ -[(h+4)^2-(h-4)^2]=32 \pm 8\sqrt{2}p~~(*) \\ \text{or,}~~ -4 \cdot h \cdot 4=32 \pm 8\sqrt{2}p \\ \text{or,}~~ -16h=32 \pm 8\sqrt{2}p \\ \text{or,}~~ -2h= 4 \pm \sqrt{2}p \\ \text{or,}~~ -(2h+4)=\pm \sqrt{2}p \\ \text{or,}~~ (2h+4)^2=2p^2 \\ \text{or,}~~ 4h^2+2 \cdot 2h \cdot 4+4^2=2(h+4)^2+2k^2 \\ \text{or,}~~ 2h^2+8h+8=(h+4)^2+k^2 \\ \text{or,}~~ 2h^2+8h+8=h^2+8h+16+k^2 \\ \therefore~h^2-k^2=8 \rightarrow(1)$

Hence, by $(1)$ we can say that the locus of moving point is $~x^2-y^2=8~$ which is a rectangular hyperbola.

Note $[*] : p=\sqrt{(h+4)^2+k^2}$

5. A point moves on a plane so that its distance from the line $~3x-4=0.~$ Show that the locus of the moving point is a hyperbola and the equation of its locus is $~5x^2-4y^2=20.$

Solution.

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

So, by question,

$\sqrt{(h-3)^2+(k-0)^2}=\frac 32 \cdot \left|\frac{3h-4}{\sqrt{3^2+0^2}}\right| \\ \text{or,}~~ \sqrt{(h-3)^2+k^2}=\frac 32 \cdot \frac{|3h-4|}{3} \\ \text{or,}~~ (h-3)^2+k^2=\frac 94 \cdot \frac{(3h-4)^2}{9} \\ \text{or,}~~ 4(h^2-6h+9)+4k^2=9h^2-24h+16 \\ \text{or,}~~ 36-16=(9h^2-4h^2)-4k^2 \\ \text{or,}~~ 5h^2-4k^2=20 \rightarrow(1)$

Hence, by $(1),~$ we can say that the locus of the moving point represents a hyperbola and is given by $~5x^2-4y^2=20.$

6. The hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point of intersection of the lines $~x-3\sqrt{5}y=0~$ and $~\sqrt{5}x-2y=13~$ and the length of its latus rectum is $~\frac 43.$ Find the co-ordinates of its foci.

Solution.

The given equations of straight lines are $~x-3\sqrt{5}y=0 \Rightarrow x=3\sqrt{5}y\rightarrow(1)~$ and $~\sqrt{5}x-2y=13 \rightarrow(2)$

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

$\sqrt{5}(3\sqrt{5}y)-2y=13 \\ \text{or,}~~ 15y-2y=13 \\ \text{or,}~~ 13y=13 \\ \text{or,}~~ y=\frac{13}{13}=1.$

$\therefore~ x=3\sqrt{5} \times 1=3\sqrt{5}.$

So, the point of intersection of the straight lines $(1)$ and $(2)$ is given by $~(3\sqrt{5},1).$

Since the given hyperbola passes through the point $~(3\sqrt{5},1),$

$\frac{(3\sqrt{5})^2}{a^2}-\frac{1^2}{b^2}=1 \\ \text{or,}~~ \frac{45}{a^2}-\frac{1}{b^2}=1 \rightarrow(3)$

Again, the length of the latus rectum of the hyperbola is

$\frac{2b^2}{a}=\frac 43 \Rightarrow b^2=\frac 23 a \rightarrow(4)$

From $(3)$ and $(4)$ we get,

$~\frac{45}{a^2}-\frac{3}{2a}=1 \\ \text{or,}~~ \frac{90-3a}{2a^2}=1 \\ \text{or,}~~ 90-3a=2a^2 \\ \text{or,}~~ 2a^2+3a-90=0 \\ \text{or,}~~ a=\frac{-3 \pm \sqrt{3^2-4 \times 2 \times (-90)}}{2 \times 2}=\frac{-3 \pm \sqrt{729}}{4} \\ \therefore~ a= \frac{-3+27}{4}=6~~(\because~ a>0)$

$\therefore~ b^2= \frac 23 \times 6=4 \Rightarrow b=2~(\because~ b>0).$

$\therefore~ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{4}{36}}=\sqrt{1+\frac 19} \\ \text{or,}~~ e=\frac{\sqrt{10}}{3}$

Hence, the co-ordinates of the foci are

$(\pm ae,0)=\left( \pm 6 \times \frac{\sqrt{10}}{3},0\right)=(\pm 2\sqrt{10},0).$

7. The numerical value of the product of the perpendicular distances of a moving point from the lines $~4x-3y+11=0~$ and $~4x+3y+5=0~$ is $~\frac{144}{25}. ~$ Find the equation to the locus of the moving point.

Solution.

Comparing the given ellipse with the general equation of the ellipse $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ we get, $~a^2=25 \Rightarrow a=\sqrt{25}=5,~~b^2=9 \Rightarrow b=\sqrt{9}=3.$

The eccentricity $(e)$ of the given ellipse is

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac 45.$

The co-ordinates of foci of the ellipse are $(\pm ae,0)=\left(\pm 5 \cdot \frac 45,0 \right)=(\pm 4,0).$

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

So, by question

$\frac{|4h-3k+11|}{\sqrt{4^2+(-3)^2}} \times \frac{|4h+3k+5|}{\sqrt{4^2+3^2}}=\frac{144}{25} \\ \text{or,}~~ \frac 15 \times \frac 15 \times |(4h-3k+11)(4h+3k+5)|=\frac{144}{25} \\ \text{or,}~~ |(4h+3k)(4h-3k)+5(4h-3k)+11(4h+3k)+55|=144 \\ \text{or,}~~ |16h^2-9k^2+20h-15k+44h+33k+55|=144 \\ \text{or,}~~ |16(h^2+4k+4)-9(k^2-2k+1)|=144 \\ \text{or,}~~|16(h+2)^2-9(k-1)^2|=144$

Hence, the locus of the moving point is given by $~16(x+2)^2-9(y-1)^2=144~$ or, $~9(y-1)^2-16(x+2)^2=144.$

8. An ellipse $E$ has the equation $~\frac{x^2}{36}+\frac{y^2}{32}=1~$ and $~C~$ and $~S~$ are its centre and focus in the usual notation. A hyperbola $~H~$ has a vertex at $C$ , has the point $S$ as its focus nearer to $C$ and has its latus rectum of equal length to that of $E.$ Show that $H$ has the eccentricity $~\frac 53~$ and find its equation.

Solution.

The given ellipse $~E: \frac{x^2}{36}+\frac{y^2}{32}=1\rightarrow(1)$

The eccentricity $(e)$ of $(1)$ is

$e=\sqrt{1-\frac{32}{36}}=\sqrt{\frac{4}{36}}=\sqrt{\frac 19}=\frac 13,\\~C\equiv(0,0),~~S \equiv (2,0).$

The transverse axis of $H$ is along $~x-$ axis and the conjugate axis of H is along $~y$ -axis. Let the eccentricity of H be $~e_1,~$ the length of the transverse axis be $~2a_1~$ and the length of the conjugate axis be $~2b_1.$

$\therefore~$ Focus of H is $~(\alpha+a_1e_1,0)~$ and the vertex $~(\alpha+a_1,0)~$ and so clearly, the ordinates of centre of $H$ is $(\beta)=0.$

By question, $~\alpha+a_1=0 \Rightarrow \alpha=-a_1~$ and

$~\alpha+a_1e_1=2 \Rightarrow -a_1+a_1e_1=2 \Rightarrow a_1(e_1-1)=2 \rightarrow(1)$

Again, the length of latus rectum of $H$ is = the length of latus rectum of $E$ .

$\therefore~\frac{2b_1^2}{a_1}=\frac{2b^2}{a}=2 \times \frac{32}{6} \\ \text{or,}~~ \frac{b_1^2}{a_1}=\frac{32}{6}=\frac{16}{3}$

For the hyperbola, the eccentricity $(e_1)$ is

$e_1=1+\frac{b_1^2}{a_1^2} \Rightarrow e_1^2-1=\frac{b_1^2}{a_1^2} \Rightarrow \frac{b_1^2}{a_1}=a_1(e_1^2-1)=\frac{16}{3} \\ \therefore~ a_1(e_1-1) \times (e_1+1)=\frac{16}{3} \\ \text{or,}~~ 2(e_1+1)=\frac{16}{3}~~[\text{By (1)}] \\ \text{or,}~~ e_1+1=\frac 83 \Rightarrow e_1=\frac 83-1=\frac 53.$

$~a_1(e_1-1)=2 \Rightarrow a_1\left(\frac 53-1\right)=2 \Rightarrow a_1 \times \frac 23=2 \\ \therefore~ a_1=3.$

$\frac{b_1^2}{a_1}=\frac{16}{3} \Rightarrow \frac{b_1^2}{3}=\frac{16}{3} \\ \therefore~ b_1^2=16 \Rightarrow b_1=\sqrt{16}=4.$

$\therefore~~\alpha=-a_1=-3,~~\beta=0.$

hence, the equation of required hyperbola is

$~\frac{(x-\alpha)^2}{a_1^2}-\frac{(y-\beta)^2}{b_1^2}=1 \\ \text{or,}~~ \frac{(x-(-3))^2}{9}-\frac{(y-0)^2}{16}=1 \\ \therefore~ \frac{(x+3)^2}{9}-\frac{y^2}{16}=1.$

9. The foci of the ellipse $~\frac{x^2}{25}+\frac{y^2}{9}=1~$ coincide with the hyperbola. If $~e=2~$ for hyperbola, find the equation of the hyperbola.

Solution.

The given equation of ellipse is $~\frac{x^2}{25}+\frac{y^2}{16}=1 \rightarrow(1)$

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

$a^2=25 \Rightarrow a=5,~~b^2=16 \Rightarrow b=4.$

The eccentricity of $(1)$ is given by

$e'=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac 45.$

The co-ordinates of foci of the ellipse

$( \pm ae',0)=\left(\pm 5 \times \frac 45,0\right)=(\pm 4,0).$

Let the equation of the hyperbola be $~\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1\rightarrow(2)$

The eccentricity of the hyperbola $~e=2~~$ . Since the foci of the ellipse $~\frac{x^2}{25}+\frac{y^2}{9}=1~$ coincide with the hyperbola $~(2)$ , so

$~a_1e=2 \Rightarrow a_1 \times 2=4 \Rightarrow a_1=\frac 42=2.$

For the hyperbola $(2)$ we get,

$~e^2=1+\frac{b_1^2}{a_1^2} \\ \text{or,}~~ 2^2=1+\frac{b_1^2}{2^2} \\ \text{or,}~~ 4-1=\frac{b_1^2}{4} \\ \text{or,}~~ b_1^2=3 \times 4=12.$

Hence, by $(2)$ , the equation of required hyperbola is

$~\frac{x^2}{4}-\frac{y^2}{12}=1 \Rightarrow 3x^2-y^2=12.$

Solution.

$~x=a \cdot \frac{1+t^2}{1-t^2},~~y=\frac{2at}{1-t^2} \\ \Rightarrow \frac xa=\frac{1+t^2}{1-t^2},~~\frac ya=\frac{2t}{1-t^2}.$

$\therefore~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=\left(\frac{1+t^2}{1-t^2}\right)^2-\left(\frac{2at}{1-t^2}\right)^2 \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=\frac{(1+t^2)^2-4t^2}{(1-t^2)^2} \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=\frac{(1-t^2)^2}{(1-t^2)^2} \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=1\rightarrow(1)$

Clearly, the equation $(1)$ represents a hyperbola whose eccentricity is given by $~e=\sqrt{1+\frac{a^2}{a^2}}=\sqrt{1+1}=\sqrt{2}.$

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

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 Complete Long Answer Type questions of Hyperbola related problems of S N Dey mathematics class 11.

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