Differentiation (Part-38) | S N De

In this article, we have solved Complete Long Answer Type questions (1-9) of Hyperbola related problems of S N Dey mathematics class 11.

Long Answer Type Questions

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,

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

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

(ii) The eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.

(iii) The co-ordinates of foci:

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

(iv) The equations of directrices are:

$$\begin{aligned} x &= \alpha \pm \frac{a}{e} = -2 \pm \frac{3}{5/3} = -2 \pm \frac{9}{5} \\ \therefore 5(x+2) &= \pm 9 \end{aligned}$$

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

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

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,

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

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

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

The eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(iii) The equations of directrices are given by

$$\begin{aligned} x &= \alpha \pm \frac{a}{e} = 1 \pm \frac{4}{5/4} = 1 \pm \frac{16}{5} \\ \text{or, } 5x &= 5 \pm 16 \\ \therefore 5x &= 21, \quad 5x = -11 \Rightarrow 5x+11=0 \end{aligned}$$

3. Find (i) the centre (ii) the vertices (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 (a)

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

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

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

(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: $2a = 2 \times 4 = 8 \text{ units}$ and the length of conjugate axis: $2b = 2 \times 3 = 6 \text{ units}$.

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

$$e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$$

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

(vii) The co-ordinates of foci:

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

(viii) The equations of directrices are given by

$$\begin{aligned} x &= \alpha \pm \frac{a}{e} = 5 \pm \frac{4}{5/4} = 5 \pm \frac{16}{5} \\ \text{or, } 5x &= 25 \pm 16 \Rightarrow 5(x-5) = \pm 16 \end{aligned}$$

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Solution (b)

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

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

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

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

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

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

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

(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: $\frac{2b^2}{a} = 2 \times \frac{13/3}{\sqrt{13/3}} = 2\sqrt{\frac{13}{3}} \text{ units}$.

(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

$$\begin{aligned} x &= \alpha \pm \frac{a}{e} = 3 \pm \frac{\sqrt{13/3}}{\sqrt{2}} = 3 \pm \sqrt{\frac{13}{6}} \\ \text{or, } x-3 &= \pm \sqrt{\frac{13}{6}} \end{aligned}$$

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 the moving point be $(h,k)$.

So, by question,

$$\begin{aligned} \left| \sqrt{(h-4)^2+k^2} - \sqrt{(h+4)^2+k^2} \right| &= 4\sqrt{2} \\ \text{or, } \sqrt{(h-4)^2+k^2} &= \sqrt{(h+4)^2+k^2} \pm 4\sqrt{2} \\ \text{or, } (h-4)^2+k^2 &= (h+4)^2+k^2 + (4\sqrt{2})^2 \pm 8\sqrt{2}\sqrt{(h+4)^2+k^2} \\ \text{or, } -[(h+4)^2-(h-4)^2] &= 32 \pm 8\sqrt{2}p \quad (*) \\ \text{or, } -4 \cdot h \cdot 4 &= 32 \pm 8\sqrt{2}p \\ \text{or, } -16h &= 32 \pm 8\sqrt{2}p \\ \text{or, } -2h &= 4 \pm \sqrt{2}p \\ \text{or, } -(2h+4) &= \pm \sqrt{2}p \\ \text{or, } (2h+4)^2 &= 2p^2 \\ \text{or, } 4h^2 + 16h + 16 &= 2[(h+4)^2+k^2] \\ \text{or, } 2h^2 + 8h + 8 &= (h+4)^2 + k^2 \\ \text{or, } 2h^2 + 8h + 8 &= h^2 + 8h + 16 + k^2 \\ \therefore h^2 - k^2 &= 8 \quad \rightarrow(1) \end{aligned}$$

Hence, by $(1)$ we can say that the locus of the 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,

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

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 $\frac{4}{3}$. Find the co-ordinates of its foci.

Solution

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

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

$$\begin{aligned} \sqrt{5}(3\sqrt{5}y) - 2y &= 13 \\ \text{or, } 15y - 2y &= 13 \\ \text{or, } 13y &= 13 \\ \text{or, } y &= \frac{13}{13} = 1 \end{aligned}$$

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

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 $(3\sqrt{5},1)$,

$$\begin{aligned} \frac{(3\sqrt{5})^2}{a^2} - \frac{1^2}{b^2} &= 1 \\ \text{or, } \frac{45}{a^2} - \frac{1}{b^2} &= 1 \quad \rightarrow(3) \end{aligned}$$

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

$$\frac{2b^2}{a} = \frac{4}{3} \Rightarrow b^2 = \frac{2}{3}a \quad \rightarrow(4)$$

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

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

$\therefore b^2 = \frac{2}{3} \times 6 = 4 \Rightarrow b=2 \quad (\because b>0)$.

$$\begin{aligned} \therefore e &= \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{4}{36}} = \sqrt{1+\frac{1}{9}} \\ \text{or, } e &= \frac{\sqrt{10}}{3} \end{aligned}$$

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

(Note: Text refers to an ellipse initially before setting up the locus equation)

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

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

$$e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{25-9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

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

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

So, by question,

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

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 $\frac{5}{3}$ and find its equation.

Solution

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

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

$$e = \sqrt{1-\frac{32}{36}} = \sqrt{\frac{4}{36}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$$

$$C \equiv (0,0), \quad S \equiv (2,0)$$

The transverse axis of $H$ is along the $x$-axis and the conjugate axis of $H$ is along the $y$-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 ordinate of centre of $H$ is $(\beta)=0$.

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

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

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

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

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

$$\begin{aligned} e_1^2 &= 1+\frac{b_1^2}{a_1^2} \Rightarrow e_1^2-1 = \frac{b_1^2}{a_1^2} \\ \Rightarrow \frac{b_1^2}{a_1} &= a_1(e_1^2-1) = \frac{16}{3} \\ \therefore a_1(e_1-1) \times (e_1+1) &= \frac{16}{3} \\ \text{or, } 2(e_1+1) &= \frac{16}{3} \quad [\text{By (2)}] \\ \text{or, } e_1+1 &= \frac{8}{3} \Rightarrow e_1 = \frac{8}{3}-1 = \frac{5}{3} \end{aligned}$$

$$a_1(e_1-1)=2 \Rightarrow a_1\left(\frac{5}{3}-1\right)=2 \Rightarrow a_1 \times \frac{2}{3}=2 \quad \therefore a_1=3$$

$$\begin{aligned} \frac{b_1^2}{a_1} &= \frac{16}{3} \Rightarrow \frac{b_1^2}{3} = \frac{16}{3} \\ \therefore b_1^2 &= 16 \Rightarrow b_1 = \sqrt{16} = 4 \end{aligned}$$

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

Hence, the equation of the required hyperbola is

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

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 $\frac{x^2}{25}+\frac{y^2}{16}=1 \quad \rightarrow(1)$ (Note: Source text has $\frac{y^2}{16}$ in solution despite $\frac{y^2}{9}$ in the question).

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

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

The eccentricity of $(1)$ is given by

$$e' = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$ (Note: Source had mathematical error leading to $4/5$)

The co-ordinates of foci of the ellipse

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

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

The eccentricity of the hyperbola $e=2$. Since the foci of the ellipse coincide with the hyperbola $(2)$, so

$$a_1e=2 \Rightarrow a_1 \times 2=4 \Rightarrow a_1=\frac{4}{2}=2$$

For the hyperbola $(2)$ we get,

$$\begin{aligned} e^2 &= 1+\frac{b_1^2}{a_1^2} \\ \text{or, } 2^2 &= 1+\frac{b_1^2}{2^2} \\ \text{or, } 4-1 &= \frac{b_1^2}{4} \\ \text{or, } b_1^2 &= 3 \times 4 = 12 \end{aligned}$$

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

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

Additional Note: Parametric Equation Proof

$$\begin{aligned} x &= a \cdot \frac{1+t^2}{1-t^2}, \quad y = \frac{2at}{1-t^2} \\ \Rightarrow \frac{x}{a} &= \frac{1+t^2}{1-t^2}, \quad \frac{y}{a} = \frac{2t}{1-t^2} \end{aligned}$$

$$\begin{aligned} \therefore \frac{x^2}{a^2} - \frac{y^2}{a^2} &= \left(\frac{1+t^2}{1-t^2}\right)^2 - \left(\frac{2t}{1-t^2}\right)^2 \\ \text{or, } \frac{x^2}{a^2} - \frac{y^2}{a^2} &= \frac{(1+t^2)^2 - 4t^2}{(1-t^2)^2} \\ \text{or, } \frac{x^2}{a^2} - \frac{y^2}{a^2} &= \frac{(1-t^2)^2}{(1-t^2)^2} \\ \text{or, } \frac{x^2}{a^2} - \frac{y^2}{a^2} &= 1 \quad \rightarrow(1) \end{aligned}$$

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