OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Exercise 25 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Ex 25(a)

 

Question 1. Find the equation of the hyperbola whose focus is (1, 1), directrix 2x + 2y = 1, and eccentricity \(\sqrt{2}\).
Answer: Let \(P(x, y)\) be any point on the hyperbola. The focus is \(F(1, 1)\), the directrix is \(2x + 2y - 1 = 0\), and the eccentricity is \(e = \sqrt{2}\). By definition of a hyperbola, the distance from \(P\) to the focus (PF) is \(e\) times the distance from \(P\) to the directrix (PM).
So, \(PF = e \cdot PM\)
The distance \(PF = \sqrt{(x-1)^2 + (y-1)^2}\)
The distance \(PM = \frac{|2x + 2y - 1|}{\sqrt{2^2 + 2^2}} = \frac{|2x + 2y - 1|}{\sqrt{4 + 4}} = \frac{|2x + 2y - 1|}{\sqrt{8}} = \frac{|2x + 2y - 1|}{2\sqrt{2}}\)
Now, substituting these into the formula \(PF = e \cdot PM\):
\( \sqrt{(x-1)^2 + (y-1)^2} = \sqrt{2} \cdot \frac{|2x + 2y - 1|}{2\sqrt{2}} \)
\( \sqrt{(x-1)^2 + (y-1)^2} = \frac{1}{2} |2x + 2y - 1| \)
To remove the square root, we square both sides:
\( (x-1)^2 + (y-1)^2 = \left( \frac{1}{2} |2x + 2y - 1| \right)^2 \)
\( (x-1)^2 + (y-1)^2 = \frac{1}{4} (2x + 2y - 1)^2 \)
Multiply both sides by 4:
\( 4[(x-1)^2 + (y-1)^2] = (2x + 2y - 1)^2 \)
Expand the terms:
\( 4[x^2 - 2x + 1 + y^2 - 2y + 1] = (2x)^2 + (2y)^2 + (-1)^2 + 2(2x)(2y) + 2(2y)(-1) + 2(-1)(2x) \)
\( 4[x^2 + y^2 - 2x - 2y + 2] = 4x^2 + 4y^2 + 1 + 8xy - 4y - 4x \)
Distribute 4 on the left side:
\( 4x^2 + 4y^2 - 8x - 8y + 8 = 4x^2 + 4y^2 + 1 + 8xy - 4y - 4x \)
Move all terms to one side to simplify:
\( (4x^2 - 4x^2) + (4y^2 - 4y^2) - 8x + 4x - 8y + 4y + 8 - 1 - 8xy = 0 \)
\( -4x - 4y + 7 - 8xy = 0 \)
Rearrange the terms to get the standard form:
\( 8xy + 4x + 4y - 7 = 0 \)
This is the required equation of the hyperbola. This formula is a powerful way to define conic sections based on their geometric properties.
In simple words: We used the special rule that says for a hyperbola, the distance from any point on it to the focus is always a fixed number of times (eccentricity) the distance to a special line called the directrix. By putting in the given focus, directrix, and eccentricity, we solved for the equation.

🎯 Exam Tip: Remember to carefully square both sides of the distance formula and expand fully. Mistakes often happen with the `\((a+b+c)^2\)` expansion or by forgetting to multiply the entire right side by `\(e^2\)`.

 

Question 2. Find the equation to the hyperbola whose eccentricity is 2, whose focus is (2, 0) and whose directrix is x - y = 0.
Answer: Let \(P(x, y)\) be any point on the hyperbola. The focus is \(F(2, 0)\), the directrix is \(x - y = 0\), and the eccentricity is \(e = 2\).
According to the definition of a hyperbola, the distance from point \(P\) to the focus \(F\) (PF) is equal to the eccentricity \(e\) multiplied by the distance from point \(P\) to the directrix (PM).
So, \(PF = e \cdot PM\)
The distance \(PF = \sqrt{(x-2)^2 + (y-0)^2} = \sqrt{(x-2)^2 + y^2}\)
The distance \(PM = \frac{|x - y|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y|}{\sqrt{1 + 1}} = \frac{|x - y|}{\sqrt{2}}\)
Substitute these values into the formula \(PF = e \cdot PM\):
\( \sqrt{(x-2)^2 + y^2} = 2 \cdot \frac{|x - y|}{\sqrt{2}} \)
\( \sqrt{(x-2)^2 + y^2} = \sqrt{2} |x - y| \)
To eliminate the square root, we square both sides of the equation:
\( (x-2)^2 + y^2 = (\sqrt{2} |x - y|)^2 \)
\( (x-2)^2 + y^2 = 2(x - y)^2 \)
Expand the terms:
\( x^2 - 4x + 4 + y^2 = 2(x^2 - 2xy + y^2) \)
\( x^2 - 4x + 4 + y^2 = 2x^2 - 4xy + 2y^2 \)
Move all terms to one side to simplify and combine like terms:
\( 2x^2 - x^2 + 2y^2 - y^2 - 4xy + 4x - 4 = 0 \)
\( x^2 + y^2 - 4xy + 4x - 4 = 0 \)
This is the required equation of the hyperbola. The eccentricity helps us understand how "stretched out" the hyperbola is.
In simple words: We used the rule that a point on the hyperbola is always a certain multiple (eccentricity) farther from its focus than from its directrix line. We set up the distances, squared both sides to get rid of square roots, and simplified to find the hyperbola's equation.

🎯 Exam Tip: When dealing with directrix equations like `\(x-y=0\)`, remember to divide by `\(\sqrt{a^2+b^2}\)` (which is `\(\sqrt{1^2+(-1)^2} = \sqrt{2}\)`) for the perpendicular distance formula.

 

Question 3. Find the equation to the conic section whose focus is (ae, 0), directrix is the line ex = a, and eccentricity is e. State whether the conic section is an ellipse or a hyperbola.
Answer: Let \(P(x, y)\) be any point on the conic section. The focus is \(F(ae, 0)\), and the directrix is the line \(ex - a = 0\). The eccentricity is \(e\).
By definition, the distance from \(P\) to the focus (PF) is \(e\) times the distance from \(P\) to the directrix (PM).
So, \(PF = e \cdot PM\)
The distance \(PF = \sqrt{(x-ae)^2 + (y-0)^2} = \sqrt{(x-ae)^2 + y^2}\)
The distance \(PM = \frac{|ex - a|}{\sqrt{e^2 + 0^2}} = \frac{|ex - a|}{e}\)
Substitute these into the formula \(PF = e \cdot PM\):
\( \sqrt{(x-ae)^2 + y^2} = e \cdot \frac{|ex - a|}{e} \)
\( \sqrt{(x-ae)^2 + y^2} = |ex - a| \)
To eliminate the square root and absolute value, we square both sides:
\( (x-ae)^2 + y^2 = (ex - a)^2 \)
Expand both sides:
\( x^2 - 2aex + a^2e^2 + y^2 = e^2x^2 - 2aex + a^2 \)
Cancel `\(-2aex\)` from both sides:
\( x^2 + a^2e^2 + y^2 = e^2x^2 + a^2 \)
Rearrange terms to group `\(x^2\)` and `\(y^2\)` terms:
\( x^2 - e^2x^2 + y^2 = a^2 - a^2e^2 \)
Factor out `\(x^2\)` and `\(a^2\)`:
\( x^2(1 - e^2) + y^2 = a^2(1 - e^2) \)
Divide by `\(a^2(1 - e^2)\)` (assuming `\(1-e^2 \neq 0\)` and `\(a^2 \neq 0\)`):
\( \frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1 \)
This is the standard equation of a conic section.
Now, let's determine the type of conic section:
If \(e < 1\), then \(1 - e^2\) is positive. In this case, `\(a^2(1-e^2)\)` is positive, and the equation represents an ellipse.
If \(e > 1\), then \(1 - e^2\) is negative. Let \(1 - e^2 = -(e^2 - 1)\). The equation becomes `\( \frac{x^2}{a^2} - \frac{y^2}{a^2(e^2 - 1)} = 1 \)`, which represents a hyperbola. The value of eccentricity determines the shape.
In simple words: We used the definition that says a point on a conic section is a fixed distance ratio (eccentricity) from a focus and a directrix. After doing the math, we got a general equation. If the eccentricity is less than 1, it's an ellipse. If it's more than 1, it's a hyperbola.

🎯 Exam Tip: Always relate the sign of `\(1-e^2\)` to determine if the conic is an ellipse or a hyperbola. If `\(e=1\)`, the denominator `\(a^2(1-e^2)\)` would be zero, leading to a parabola.

 

Question 4. Find the equation of the hyperbola whose axes are along the coordinate axes and which passes through (- 3, 4), and (5, 6).
Answer: A hyperbola with axes along the coordinate axes has an equation of the form `\(\frac{x^2}{A} - \frac{y^2}{B} = 1\)` or `\(\frac{y^2}{A} - \frac{x^2}{B} = 1\)`, where \(A\) and \(B\) are positive. We will try both forms.
First, let's assume the equation is `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)` (transverse axis on the x-axis).
Since the hyperbola passes through `\((-3, 4)\)`:
`\( \frac{(-3)^2}{a^2} - \frac{4^2}{b^2} = 1 \implies \frac{9}{a^2} - \frac{16}{b^2} = 1 \quad \text{...(1)} \)`
Since the hyperbola passes through `\((5, 6)\)`:
`\( \frac{5^2}{a^2} - \frac{6^2}{b^2} = 1 \implies \frac{25}{a^2} - \frac{36}{b^2} = 1 \quad \text{...(2)} \)`
Let \(X = \frac{1}{a^2}\) and \(Y = \frac{1}{b^2}\). The equations become:
`\( 9X - 16Y = 1 \quad \text{...(1')} \)`
`\( 25X - 36Y = 1 \quad \text{...(2')} \)`
Multiply (1') by 25 and (2') by 9:
`\( 225X - 400Y = 25 \)`
`\( 225X - 324Y = 9 \)`
Subtract the second new equation from the first:
`\( (-400Y) - (-324Y) = 25 - 9 \)`
`\( -76Y = 16 \implies Y = -\frac{16}{76} = -\frac{4}{19} \)`
Since \(Y = \frac{1}{b^2}\), a negative value means `\(b^2\)` would be negative, which is not possible for a hyperbola in this form. This tells us our initial assumption for the equation was incorrect.

Now, let's assume the equation is `\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)` (transverse axis on the y-axis).
Since the hyperbola passes through `\((-3, 4)\)`:
`\( \frac{4^2}{a^2} - \frac{(-3)^2}{b^2} = 1 \implies \frac{16}{a^2} - \frac{9}{b^2} = 1 \quad \text{...(3)} \)`
Since the hyperbola passes through `\((5, 6)\)`:
`\( \frac{6^2}{a^2} - \frac{5^2}{b^2} = 1 \implies \frac{36}{a^2} - \frac{25}{b^2} = 1 \quad \text{...(4)} \)`
Let \(X = \frac{1}{a^2}\) and \(Y = \frac{1}{b^2}\). The equations become:
`\( 16X - 9Y = 1 \quad \text{...(3')} \)`
`\( 36X - 25Y = 1 \quad \text{...(4')} \)`
Multiply (3') by 25 and (4') by 9:
`\( 400X - 225Y = 25 \)`
`\( 324X - 225Y = 9 \)`
Subtract the second new equation from the first:
`\( (400X) - (324X) = 25 - 9 \)`
`\( 76X = 16 \implies X = \frac{16}{76} = \frac{4}{19} \)`
Since \(X = \frac{1}{a^2}\), we have `\( \frac{1}{a^2} = \frac{4}{19} \implies a^2 = \frac{19}{4} \)`
Substitute \(X = \frac{4}{19}\) into (3'):
`\( 16 \left(\frac{4}{19}\right) - 9Y = 1 \)`
`\( \frac{64}{19} - 9Y = 1 \)`
`\( 9Y = \frac{64}{19} - 1 = \frac{64 - 19}{19} = \frac{45}{19} \)`
`\( Y = \frac{45}{19 \times 9} = \frac{5}{19} \)`
Since \(Y = \frac{1}{b^2}\), we have `\( \frac{1}{b^2} = \frac{5}{19} \implies b^2 = \frac{19}{5} \)`
Now substitute `\(a^2\)` and `\(b^2\)` back into the hyperbola equation `\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)`:
`\( \frac{y^2}{(19/4)} - \frac{x^2}{(19/5)} = 1 \)`
\( \frac{4y^2}{19} - \frac{5x^2}{19} = 1 \)
Multiply the entire equation by 19:
\( 4y^2 - 5x^2 = 19 \)
This is the required equation of the hyperbola. This approach of checking both axis orientations is crucial when points are given.
In simple words: We used the general equation for a hyperbola aligned with the axes. We plugged in the two given points to create two equations. Since solving them for the x-axis alignment didn't work (giving a negative square), we tried the y-axis alignment. That gave us valid numbers for \(a^2\) and \(b^2\), which we then used to write the final equation.

🎯 Exam Tip: When the axes are along the coordinate axes and points are given, always consider both possible standard forms for the hyperbola `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)` and `\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)` if not specified. One will lead to positive `\(a^2\)` and `\(b^2\)` values, the other won't.

 

Question 5. Find the eccentricity of the hyperbola whose equation is 2x² – 3y² = 15.
Answer: The given equation of the hyperbola is `\(2x^2 - 3y^2 = 15\)`.
To find the eccentricity, we first need to convert the equation into the standard form of a hyperbola, which is `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)` or `\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)`.
Divide the entire equation by 15:
`\( \frac{2x^2}{15} - \frac{3y^2}{15} = \frac{15}{15} \)`
`\( \frac{x^2}{15/2} - \frac{y^2}{5} = 1 \)`
Now, we compare this equation with the standard form `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)`:
We can see that `\(a^2 = \frac{15}{2}\)` and `\(b^2 = 5\)`.
The relationship between `\(a^2\), \(b^2\)`, and the eccentricity `\(e\)` for a hyperbola with transverse axis along the x-axis is `\(b^2 = a^2(e^2 - 1)\)`.
Substitute the values of `\(a^2\)` and `\(b^2\)`:
\( 5 = \frac{15}{2} (e^2 - 1) \)
Divide both sides by `\(\frac{15}{2}\)`:
\( e^2 - 1 = \frac{5 \times 2}{15} = \frac{10}{15} = \frac{2}{3} \)
Now, solve for `\(e^2\)`:
\( e^2 = 1 + \frac{2}{3} = \frac{3 + 2}{3} = \frac{5}{3} \)
Finally, find `\(e\)`:
\( e = \sqrt{\frac{5}{3}} \)
Since eccentricity `\(e\)` must be positive for a hyperbola, we take the positive square root. The eccentricity tells us how wide the hyperbola opens.
In simple words: First, we changed the given hyperbola equation into its standard form to find the values of \(a^2\) and \(b^2\). Then, we used the special formula that connects \(a^2\), \(b^2\), and eccentricity for a hyperbola to solve for \(e\).

🎯 Exam Tip: Always remember to convert the given equation to standard form before identifying `\(a^2\)` and `\(b^2\)`. Use the correct eccentricity formula: `\(b^2 = a^2(e^2-1)\)` for horizontal transverse axis and `\(a^2 = b^2(e^2-1)\)` for vertical transverse axis.

 

Question 6. Find the eccentricity and the coordinates of the foci of the curve 3x² – y² = 4.
Answer: The given equation of the curve is `\(3x^2 - y^2 = 4\)`.
To find the eccentricity and foci, we convert the equation to the standard form of a hyperbola: `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)`.
Divide the entire equation by 4:
`\( \frac{3x^2}{4} - \frac{y^2}{4} = \frac{4}{4} \)`
`\( \frac{x^2}{4/3} - \frac{y^2}{4} = 1 \)`
Comparing this with the standard form `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)`:
We have `\(a^2 = \frac{4}{3}\)` and `\(b^2 = 4\)`.
Now, let's find the eccentricity `\(e\)` using the formula `\(b^2 = a^2(e^2 - 1)\)`:
\( 4 = \frac{4}{3} (e^2 - 1) \)
Divide by `\(\frac{4}{3}\)`:
\( e^2 - 1 = \frac{4 \times 3}{4} = 3 \)
\( e^2 = 1 + 3 = 4 \)
So, `\(e = \sqrt{4} = 2\)` (since eccentricity `\(e\)` is always positive for a hyperbola).
The eccentricity of the hyperbola is 2.

Next, let's find the coordinates of the foci. For a hyperbola in this form (transverse axis along the x-axis), the foci are at `\((\pm ae, 0)\)`.
First, find `\(a\)` from `\(a^2 = \frac{4}{3}\)`:
\( a = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \)
Now, calculate `\(ae\)`:
\( ae = \frac{2}{\sqrt{3}} \times 2 = \frac{4}{\sqrt{3}} \)
So, the coordinates of the foci are `\( \left(\pm \frac{4}{\sqrt{3}}, 0\right) \)`. Foci are the two fixed points that define the hyperbola.
In simple words: We first rewrote the equation into a standard form to easily pick out \(a^2\) and \(b^2\). Then, we used a formula involving \(a^2\), \(b^2\), and \(e\) to find the eccentricity. Finally, we used the values of \(a\) and \(e\) to locate the two focus points of the hyperbola.

🎯 Exam Tip: Remember that for `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)`, the foci are `\((\pm ae, 0)\)`. If the `\(y^2\)` term comes first (transverse axis vertical), the foci would be `\((0, \pm ae)\)`. Ensure `\(e\)` is always positive.

 

Question 7. Find the coordinates of the foci, vertices, eccentricity and the length of the latus rectum of the hyperbola
(i) 16x² – 9y² = 576
(ii) \( \frac{y^2}{9} – \frac{x^2}{27} = 1 \)
(iii) 9y² – 4x² = 36
(iv) 49y² – 16x² = 784
Answer:
(i) Given equation: `\(16x^2 - 9y^2 = 576\)`
To convert to standard form, divide by 576:
`\( \frac{16x^2}{576} - \frac{9y^2}{576} = 1 \)`
`\( \frac{x^2}{36} - \frac{y^2}{64} = 1 \)`
Comparing with `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)`:
`\(a^2 = 36 \implies a = 6\)`
`\(b^2 = 64 \implies b = 8\)`

1. **Eccentricity (\(e\))**: Use `\(b^2 = a^2(e^2 - 1)\)`
\( 64 = 36(e^2 - 1) \)
\( e^2 - 1 = \frac{64}{36} = \frac{16}{9} \)
\( e^2 = 1 + \frac{16}{9} = \frac{25}{9} \)
\( e = \sqrt{\frac{25}{9}} = \frac{5}{3} \)

2. **Foci**: `\((\pm ae, 0)\)`
\( ae = 6 \times \frac{5}{3} = 10 \)
Coordinates of foci are `\((\pm 10, 0)\)`.

3. **Vertices**: `\((\pm a, 0)\)`
Coordinates of vertices are `\((\pm 6, 0)\)`.

4. **Length of transverse axis**: `\(2a = 2 \times 6 = 12\)`

5. **Length of conjugate axis**: `\(2b = 2 \times 8 = 16\)`

6. **Length of latus rectum**: `\(\frac{2b^2}{a} = \frac{2 \times 64}{6} = \frac{64}{3}\)

(ii) Given equation: `\( \frac{y^2}{9} - \frac{x^2}{27} = 1 \)`
This is already in standard form, with the transverse axis along the y-axis. Comparing with `\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)`:
`\(a^2 = 9 \implies a = 3\)`
`\(b^2 = 27 \implies b = \sqrt{27} = 3\sqrt{3}\)`

1. **Eccentricity (\(e\))**: Use `\(b^2 = a^2(e^2 - 1)\)`
\( 27 = 9(e^2 - 1) \)
\( e^2 - 1 = \frac{27}{9} = 3 \)
\( e^2 = 1 + 3 = 4 \)
\( e = \sqrt{4} = 2 \)

2. **Foci**: `\((0, \pm ae)\)`
\( ae = 3 \times 2 = 6 \)
Coordinates of foci are `\((0, \pm 6)\)`.

3. **Vertices**: `\((0, \pm a)\)`
Coordinates of vertices are `\((0, \pm 3)\)`.

4. **Length of latus rectum**: `\(\frac{2b^2}{a} = \frac{2 \times 27}{3} = 18\)`

(iii) Given equation: `\(9y^2 - 4x^2 = 36\)`
To convert to standard form, divide by 36:
`\( \frac{9y^2}{36} - \frac{4x^2}{36} = 1 \)`
`\( \frac{y^2}{4} - \frac{x^2}{9} = 1 \)`
Comparing with `\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)`:
`\(a^2 = 4 \implies a = 2\)`
`\(b^2 = 9 \implies b = 3\)`

1. **Eccentricity (\(e\))**: Use `\(b^2 = a^2(e^2 - 1)\)`
\( 9 = 4(e^2 - 1) \)
\( e^2 - 1 = \frac{9}{4} \)
\( e^2 = 1 + \frac{9}{4} = \frac{13}{4} \)
\( e = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \)

2. **Foci**: `\((0, \pm ae)\)`
\( ae = 2 \times \frac{\sqrt{13}}{2} = \sqrt{13} \)
Coordinates of foci are `\((0, \pm \sqrt{13})\)`.

3. **Vertices**: `\((0, \pm a)\)`
Coordinates of vertices are `\((0, \pm 2)\)`.

4. **Length of transverse axis**: `\(2a = 2 \times 2 = 4\)`

5. **Length of conjugate axis**: `\(2b = 2 \times 3 = 6\)`

6. **Length of latus rectum**: `\(\frac{2b^2}{a} = \frac{2 \times 9}{2} = 9\)`

(iv) Given equation: `\(49y^2 - 16x^2 = 784\)`
To convert to standard form, divide by 784:
`\( \frac{49y^2}{784} - \frac{16x^2}{784} = 1 \)`
`\( \frac{y^2}{16} - \frac{x^2}{49} = 1 \)`
Comparing with `\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)`:
`\(a^2 = 16 \implies a = 4\)`
`\(b^2 = 49 \implies b = 7\)`

1. **Eccentricity (\(e\))**: Use `\(b^2 = a^2(e^2 - 1)\)`
\( 49 = 16(e^2 - 1) \)
\( e^2 - 1 = \frac{49}{16} \)
\( e^2 = 1 + \frac{49}{16} = \frac{65}{16} \)
\( e = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4} \)

2. **Foci**: `\((0, \pm ae)\)`
\( ae = 4 \times \frac{\sqrt{65}}{4} = \sqrt{65} \)
Coordinates of foci are `\((0, \pm \sqrt{65})\)`.

3. **Vertices**: `\((0, \pm a)\)`
Coordinates of vertices are `\((0, \pm 4)\)`.

4. **Length of transverse axis**: `\(2a = 2 \times 4 = 8\)`

5. **Length of conjugate axis**: `\(2b = 2 \times 7 = 14\)`

6. **Length of latus rectum**: `\(\frac{2b^2}{a} = \frac{2 \times 49}{4} = \frac{49}{2}\)`
In each case, we carefully found the key features of the hyperbola by first converting its equation to a standard form and then applying specific formulas. The position of the transverse axis (horizontal or vertical) is determined by which term comes first in the standard form.
In simple words: For each hyperbola, we first changed its equation into a simple standard form. Then, using formulas, we found its eccentricity (how wide it opens), the locations of its foci (special points), its vertices (endpoints of the main axis), and the length of its latus rectum (a chord through the focus). We noted if the main axis was horizontal or vertical based on the equation.

🎯 Exam Tip: Pay close attention to whether the `\(x^2\)` term or `\(y^2\)` term is positive in the standard form. This determines if the transverse axis is horizontal or vertical, which in turn affects the coordinates of the foci and vertices.

 

Question 8. In the hyperbola x² – 4y² = 4, find the length of the axes, the coordinates of the foci, the eccentricity, and the latus rectum, and the equations of the directrices.
Answer: The given equation of the hyperbola is `\(x^2 - 4y^2 = 4\)`.
To find all the required properties, we first convert the equation to its standard form, `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)`:
Divide the entire equation by 4:
`\( \frac{x^2}{4} - \frac{4y^2}{4} = \frac{4}{4} \)`
`\( \frac{x^2}{4} - \frac{y^2}{1} = 1 \)`
Comparing this with `\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)`:
We get `\(a^2 = 4 \implies a = 2\)`
And `\(b^2 = 1 \implies b = 1\)`

1. **Length of the axes**: * **Transverse axis**: `\(2a = 2 \times 2 = 4\)` * **Conjugate axis**: `\(2b = 2 \times 1 = 2\)`

2. **Eccentricity (\(e\))**: Use the formula `\(b^2 = a^2(e^2 - 1)\)`
\( 1 = 4(e^2 - 1) \)
\( e^2 - 1 = \frac{1}{4} \)
\( e^2 = 1 + \frac{1}{4} = \frac{5}{4} \)
\( e = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \)

3. **Coordinates of the foci**: For this form, foci are `\((\pm ae, 0)\)`
\( ae = 2 \times \frac{\sqrt{5}}{2} = \sqrt{5} \)
The coordinates of the foci are `\((\pm \sqrt{5}, 0)\)`.

4. **Length of the latus rectum**: `\(\frac{2b^2}{a}\)`
\( \text{Length of latus rectum} = \frac{2 \times 1}{2} = 1 \)

5. **Equations of the directrices**: For this form, directrices are `\(x = \pm \frac{a}{e}\)`
\( x = \pm \frac{2}{(\sqrt{5}/2)} = \pm \frac{4}{\sqrt{5}} \)
The two equations of the directrices are `\(x = \frac{4}{\sqrt{5}}\)` and `\(x = -\frac{4}{\sqrt{5}}\)`. These can also be written as `\(\sqrt{5}x - 4 = 0\)` and `\(\sqrt{5}x + 4 = 0\)`. Each part of the hyperbola equation helps define its shape and position accurately.
In simple words: We converted the hyperbola's equation to its standard form. From this, we found the lengths of its two axes, how much it spreads out (eccentricity), the exact locations of its two special focus points, the length of a specific chord (latus rectum), and the equations for the two defining directrix lines.

🎯 Exam Tip: Remember all five key properties (axes lengths, eccentricity, foci, latus rectum, directrices) for both horizontal and vertical transverse hyperbolas. Memorizing the standard forms and corresponding formulas is essential.

 

Question 9. Find (a) the eccentricities, (b) the coordinates of the foci (c) the equations of the directrices of the following hyperbolas
(i) \( \frac{(x-1)^2}{9} – \frac{y^2}{4} = 1 \)
(ii) \( \frac{(x+1)^2}{64} – \frac{(y-2)^2}{36} = 1 \)
Answer:
(i) Given equation: `\( \frac{(x-1)^2}{9} - \frac{y^2}{4} = 1 \)`
This equation represents a hyperbola whose center is not at the origin. We can shift the origin to \( (1, 0) \) by letting \(X = x - 1\) and \(Y = y\).
The new equation becomes: `\( \frac{X^2}{9} - \frac{Y^2}{4} = 1 \)`
Comparing this with the standard form `\( \frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1 \)`:
`\( a^2 = 9 \implies a = 3 \)`
`\( b^2 = 4 \implies b = 2 \)`

(a) **Eccentricity (\(e\))**: Use `\(b^2 = a^2(e^2 - 1)\)`
\( 4 = 9(e^2 - 1) \)
\( e^2 - 1 = \frac{4}{9} \)
\( e^2 = 1 + \frac{4}{9} = \frac{13}{9} \)
\( e = \frac{\sqrt{13}}{3} \)

(b) **Coordinates of the foci**: For the transformed coordinates `\((X, Y)\)`, the foci are `\((\pm ae, 0)\)`
\( ae = 3 \times \frac{\sqrt{13}}{3} = \sqrt{13} \)
So, `\(X = \pm \sqrt{13}\)` and `\(Y = 0\)`.
Now, convert back to `\((x, y)\)` coordinates using `\(x - 1 = X\)` and `\(y = Y\)`:
`\( x - 1 = \pm \sqrt{13} \implies x = 1 \pm \sqrt{13} \)`
`\( y = 0 \)`
The coordinates of the foci are `\( (1 \pm \sqrt{13}, 0) \)`.

(c) **Equations of the directrices**: For the transformed coordinates `\((X, Y)\)`, the directrices are `\(X = \pm \frac{a}{e}\)`
\( X = \pm \frac{3}{(\sqrt{13}/3)} = \pm \frac{9}{\sqrt{13}} \)
Convert back to `\((x, y)\)` coordinates:
`\( x - 1 = \pm \frac{9}{\sqrt{13}} \implies x = 1 \pm \frac{9}{\sqrt{13}} \)`
The equations of the directrices are `\( x = 1 + \frac{9}{\sqrt{13}} \)` and `\( x = 1 - \frac{9}{\sqrt{13}} \)`.

(ii) Given equation: `\( \frac{(x+1)^2}{64} - \frac{(y-2)^2}{36} = 1 \)`
This hyperbola is centered at `\((-1, 2)\)`. We shift the origin by letting `\(X = x + 1\)` and `\(Y = y - 2\).
The new equation becomes: `\( \frac{X^2}{64} - \frac{Y^2}{36} = 1 \)`
Comparing this with the standard form `\( \frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1 \)`:
`\( a^2 = 64 \implies a = 8 \)`
`\( b^2 = 36 \implies b = 6 \)`

(a) **Eccentricity (\(e\))**: Use `\(b^2 = a^2(e^2 - 1)\)`
\( 36 = 64(e^2 - 1) \)
\( e^2 - 1 = \frac{36}{64} = \frac{9}{16} \)
\( e^2 = 1 + \frac{9}{16} = \frac{25}{16} \)
\( e = \sqrt{\frac{25}{16}} = \frac{5}{4} \)

(b) **Coordinates of the foci**: For the transformed coordinates `\((X, Y)\)`, the foci are `\((\pm ae, 0)\)`
\( ae = 8 \times \frac{5}{4} = 10 \)
So, `\(X = \pm 10\)` and `\(Y = 0\)`.
Convert back to `\((x, y)\)` coordinates using `\(x + 1 = X\)` and `\(y - 2 = Y\)`:
`\( x + 1 = \pm 10 \implies x = -1 \pm 10 \)`
This gives `\(x = 9\)` or `\(x = -11\)`.
`\( y - 2 = 0 \implies y = 2 \)`
The coordinates of the foci are `\( (9, 2) \)` and `\( (-11, 2) \)`.

(c) **Equations of the directrices**: For the transformed coordinates `\((X, Y)\)`, the directrices are `\(X = \pm \frac{a}{e}\)`
\( X = \pm \frac{8}{(5/4)} = \pm \frac{32}{5} \)
Convert back to `\((x, y)\)` coordinates:
`\( x + 1 = \pm \frac{32}{5} \implies x = -1 \pm \frac{32}{5} \)`
This gives two directrices:
`\( x = -1 + \frac{32}{5} = \frac{-5 + 32}{5} = \frac{27}{5} \)`
`\( x = -1 - \frac{32}{5} = \frac{-5 - 32}{5} = -\frac{37}{5} \)`
The equations of the directrices are `\( 5x - 27 = 0 \)` and `\( 5x + 37 = 0 \)`.
When the center of the hyperbola is not at the origin, we use a coordinate transformation to simplify the problem, solve it in the new coordinate system, and then transform the results back. This method makes calculations more manageable.
In simple words: For each hyperbola, we first moved the center to the origin mathematically using new \(X\) and \(Y\) coordinates. Then, we found the eccentricity, the locations of the foci, and the equations of the directrix lines using standard formulas. Finally, we changed these answers back to the original \(x\) and \(y\) coordinates.

🎯 Exam Tip: When the hyperbola's center is shifted, remember to apply the coordinate transformation `\(X=x-h\)` and `\(Y=y-k\)` consistently. Calculate all parameters in the `\((X, Y)\)` system first, then convert back to `\((x, y)\)` for the final answer. Double-check your signs when shifting coordinates.

 

Question 10. Find the equation of the hyperbola, referred to its axes as the axes of coordinates,
(i) whose transverse and conjugate axes are in length respectively 2 and 3;
(ii) whose foci are (2, 0) and (- 2, 0) and eccentricity equal to \( \frac{3}{2} \);
(iii) the distance between whose foci is 4 and whose eccentricity is \( \sqrt{2} \);
(iv) whose conjugate axis is 3 and the distance between whose foci is 5 ;
(v) vertices are (0, ± 3) and foci (0, ± 5);
(vi) foci are (± 5, 0) and transverse axis of length 8;
(vii) foci are (0, ± 13) and conjugate axis of length 24 ;
(viii) Since the vertices (± 7, 0) of required hyperbola are lies on x-axis.
(ix) foci are (6, 4) and (- 4, 4) and eccentricity is 2.
Answer:
(i) **Transverse and conjugate axes lengths are 2 and 3 respectively.**
Given length of transverse axis `\(2a = 2 \implies a = 1\)`. So, `\(a^2 = 1\)`.
Given length of conjugate axis `\(2b = 3 \implies b = \frac{3}{2}\)`. So, `\(b^2 = \frac{9}{4}\)`.
Since the transverse axis has length 2, we assume it's along the x-axis. The equation of the hyperbola is `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)`
Substitute `\(a^2\)` and `\(b^2\)`:
`\( \frac{x^2}{1} - \frac{y^2}{9/4} = 1 \)`
`\( x^2 - \frac{4y^2}{9} = 1 \)`
Multiply by 9:
`\( 9x^2 - 4y^2 = 9 \)`
This is the required equation of the hyperbola. The lengths of the axes directly provide `\(a\)` and `\(b\)` values.

(ii) **Foci are (2, 0) and (- 2, 0) and eccentricity \(e = \frac{3}{2}\).**
Since the foci are `\((\pm 2, 0)\)`, the transverse axis is along the x-axis. The center is at the origin.
From the foci, `\(ae = 2\)`.
Given `\(e = \frac{3}{2}\)`.
So, `\(a \left(\frac{3}{2}\right) = 2 \implies a = \frac{4}{3}\)`. Thus, `\(a^2 = \frac{16}{9}\)`.
Now find `\(b^2\)` using `\(b^2 = a^2(e^2 - 1)\)`:
`\( b^2 = \frac{16}{9} \left( \left(\frac{3}{2}\right)^2 - 1 \right) \)`
`\( b^2 = \frac{16}{9} \left( \frac{9}{4} - 1 \right) \)`
`\( b^2 = \frac{16}{9} \left( \frac{9 - 4}{4} \right) = \frac{16}{9} \times \frac{5}{4} = \frac{4 \times 5}{9} = \frac{20}{9} \)`
The equation of the hyperbola is `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)`
`\( \frac{x^2}{16/9} - \frac{y^2}{20/9} = 1 \)`
`\( \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \)`
To clear denominators, multiply by the LCM of 16 and 20, which is 80:
`\( 5(9x^2) - 4(9y^2) = 80 \)`
`\( 45x^2 - 36y^2 = 80 \)`
This is the required equation. The foci and eccentricity provide enough information to define the hyperbola's parameters.

(iii) **Distance between foci is 4 and eccentricity \(e = \sqrt{2}\).**
Given distance between foci `\(2ae = 4 \implies ae = 2\)`.
Given eccentricity `\(e = \sqrt{2}\)`.
Substitute `\(e\)` into `\(ae = 2\)`:
`\( a\sqrt{2} = 2 \implies a = \frac{2}{\sqrt{2}} = \sqrt{2}\)`. Thus, `\(a^2 = 2\)`.
Now find `\(b^2\)` using `\(b^2 = a^2(e^2 - 1)\)`:
`\( b^2 = 2 ((\sqrt{2})^2 - 1) \)`
`\( b^2 = 2 (2 - 1) = 2 \times 1 = 2 \)`
The equation of the hyperbola is `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)` (assuming transverse axis along x-axis)
`\( \frac{x^2}{2} - \frac{y^2}{2} = 1 \)`
Multiply by 2:
`\( x^2 - y^2 = 2 \)`
This is the required equation. The relationship `\(b^2 = a^2(e^2-1)\)` is key for finding `\(b^2\)`.

(iv) **Conjugate axis is 3 and the distance between whose foci is 5.**
Given length of conjugate axis `\(2b = 3 \implies b = \frac{3}{2}\)`. So, `\(b^2 = \frac{9}{4}\)`.
Given distance between foci `\(2ae = 5 \implies ae = \frac{5}{2}\)`. So, `\((ae)^2 = \frac{25}{4}\)`.
We know that `\(b^2 = a^2(e^2 - 1)\)`, which can be rewritten as `\(b^2 = a^2e^2 - a^2 = (ae)^2 - a^2\)`
Substitute the known values:
`\( \frac{9}{4} = \frac{25}{4} - a^2 \)`
`\( a^2 = \frac{25}{4} - \frac{9}{4} = \frac{16}{4} = 4 \)`
The equation of the hyperbola is `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)` (assuming transverse axis along x-axis)
`\( \frac{x^2}{4} - \frac{y^2}{9/4} = 1 \)`
`\( \frac{x^2}{4} - \frac{4y^2}{9} = 1 \)`
To clear denominators, multiply by the LCM of 4 and 9, which is 36:
`\( 9x^2 - 16y^2 = 36 \)`
This is the required equation. The relationships between `\(a, b, e\)` are central to solving such problems.

(v) **Vertices are (0, ± 3) and foci (0, ± 5).**
Since the vertices are `\((0, \pm 3)\)`, the transverse axis is along the y-axis. So, `\(a = 3\)` and `\(a^2 = 9\)`.
Since the foci are `\((0, \pm 5)\)`, we have `\(ae = 5\)`.
Substitute `\(a = 3\)` into `\(ae = 5\)`:
`\( 3e = 5 \implies e = \frac{5}{3} \)`
Now find `\(b^2\)` using `\(b^2 = a^2(e^2 - 1)\)`:
`\( b^2 = 9 \left( \left(\frac{5}{3}\right)^2 - 1 \right) \)`
`\( b^2 = 9 \left( \frac{25}{9} - 1 \right) \)`
`\( b^2 = 9 \left( \frac{25 - 9}{9} \right) = 9 \times \frac{16}{9} = 16 \)`
The equation of the hyperbola is `\( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)`
`\( \frac{y^2}{9} - \frac{x^2}{16} = 1 \)`
This is the required equation. The orientation of the transverse axis depends on the location of the vertices and foci.

(vi) **Foci are (± 5, 0) and transverse axis of length 8.**
Since the foci are `\((\pm 5, 0)\)`, the transverse axis is along the x-axis. So, `\(ae = 5\)`.
Given length of transverse axis `\(2a = 8 \implies a = 4\)`. So, `\(a^2 = 16\)`.
Substitute `\(a = 4\)` into `\(ae = 5\)`:
`\( 4e = 5 \implies e = \frac{5}{4} \)`
Now find `\(b^2\)` using `\(b^2 = a^2(e^2 - 1)\)`:
`\( b^2 = 16 \left( \left(\frac{5}{4}\right)^2 - 1 \right) \)`
`\( b^2 = 16 \left( \frac{25}{16} - 1 \right) \)`
`\( b^2 = 16 \left( \frac{25 - 16}{16} \right) = 16 \times \frac{9}{16} = 9 \)`
The equation of the hyperbola is `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)`
`\( \frac{x^2}{16} - \frac{y^2}{9} = 1 \)`
This is the required equation. Knowing the foci and transverse axis length directly gives `\(a\)` and `\(e\)`, allowing `\(b\)` to be found.

(vii) **Foci are (0, ± 13) and conjugate axis of length 24.**
Since the foci are `\((0, \pm 13)\)`, the transverse axis is along the y-axis. So, `\(ae = 13\)`.
Given length of conjugate axis `\(2b = 24 \implies b = 12\)`. So, `\(b^2 = 144\)`.
We use the relationship `\(b^2 = (ae)^2 - a^2\)` (derived from `\(b^2 = a^2(e^2-1)\)`).
`\( 144 = 13^2 - a^2 \)`
`\( 144 = 169 - a^2 \)`
`\( a^2 = 169 - 144 = 25 \)`
The equation of the hyperbola is `\( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)`
`\( \frac{y^2}{25} - \frac{x^2}{144} = 1 \)`
To clear denominators, multiply by `\(25 \times 144 = 3600\)`:
`\( 144y^2 - 25x^2 = 3600 \)`
This is the required equation. When the foci are on the y-axis, the standard form of the hyperbola flips roles for \(x\) and \(y\).

(viii) **Vertices are (± 7, 0) and eccentricity \(e = \frac{4}{3}\).**
Since the vertices are `\((\pm 7, 0)\)`, the transverse axis is along the x-axis. So, `\(a = 7\)` and `\(a^2 = 49\)`.
Given eccentricity `\(e = \frac{4}{3}\)`.
Now find `\(b^2\)` using `\(b^2 = a^2(e^2 - 1)\)`:
`\( b^2 = 49 \left( \left(\frac{4}{3}\right)^2 - 1 \right) \)`
`\( b^2 = 49 \left( \frac{16}{9} - 1 \right) \)`
`\( b^2 = 49 \left( \frac{16 - 9}{9} \right) = 49 \times \frac{7}{9} = \frac{343}{9} \)`
The equation of the hyperbola is `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)`
`\( \frac{x^2}{49} - \frac{y^2}{343/9} = 1 \)`
`\( \frac{x^2}{49} - \frac{9y^2}{343} = 1 \)`
To clear denominators, multiply by 343:
`\( 7x^2 - 9y^2 = 343 \)`
This is the required equation. The position of the vertices determines the axis orientation.

(ix) **Foci are (6, 4) and (- 4, 4) and eccentricity is 2.**
This hyperbola is not centered at the origin. First, find the center `\(C(\alpha, \beta)\)`, which is the midpoint of the foci:
`\( \alpha = \frac{6 + (-4)}{2} = \frac{2}{2} = 1 \)`
`\( \beta = \frac{4 + 4}{2} = \frac{8}{2} = 4 \)`
So, the center of the hyperbola is `\( (1, 4) \)`.
Since the y-coordinates of the foci are the same (4), the transverse axis is parallel to the x-axis.
The distance between the foci is `\(2ae = \sqrt{(6 - (-4))^2 + (4 - 4)^2} = \sqrt{10^2 + 0^2} = 10 \)`
So, `\(2ae = 10 \implies ae = 5\)`.
Given eccentricity `\(e = 2\)`.
Substitute `\(e = 2\)` into `\(ae = 5\)`:
`\( a(2) = 5 \implies a = \frac{5}{2}\)`. Thus, `\(a^2 = \frac{25}{4}\)`.
Now find `\(b^2\)` using `\(b^2 = a^2(e^2 - 1)\)`:
`\( b^2 = \frac{25}{4} (2^2 - 1) \)`
`\( b^2 = \frac{25}{4} (4 - 1) = \frac{25}{4} \times 3 = \frac{75}{4} \)`
The equation of the hyperbola with center `\( (h, k) \)` and transverse axis parallel to the x-axis is `\( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)`
Substitute `\(h=1, k=4, a^2=\frac{25}{4}, b^2=\frac{75}{4}\)`:
`\( \frac{(x-1)^2}{25/4} - \frac{(y-4)^2}{75/4} = 1 \)`
`\( \frac{4(x-1)^2}{25} - \frac{4(y-4)^2}{75} = 1 \)`
To clear denominators, multiply by the LCM of 25 and 75, which is 75:
`\( 3 \times 4(x-1)^2 - 4(y-4)^2 = 75 \)`
`\( 12(x-1)^2 - 4(y-4)^2 = 75 \)`
Expand the terms:
`\( 12(x^2 - 2x + 1) - 4(y^2 - 8y + 16) = 75 \)`
`\( 12x^2 - 24x + 12 - 4y^2 + 32y - 64 = 75 \)`
Combine like terms and move 75 to the left side:
`\( 12x^2 - 4y^2 - 24x + 32y + 12 - 64 - 75 = 0 \)`
`\( 12x^2 - 4y^2 - 24x + 32y - 127 = 0 \)`
This is the required equation. Problems with shifted centers require careful tracking of coordinates.
In simple words: For each part, we used the given information like axis lengths, foci, vertices, or eccentricity to find the values of \(a^2\) and \(b^2\). We also figured out if the hyperbola's main axis was horizontal or vertical, or if its center was shifted. Then we plugged these values into the correct standard equation of a hyperbola to get the final answer.

🎯 Exam Tip: For problems where the center is not at the origin, always first find the center coordinates. Then, treat it as a standard hyperbola problem by shifting the origin, calculating parameters, and finally shifting back to the original coordinates.

 

Question 11. Find the equation of the hyperbola, referred to its axes as the axes of co- ordinates, whose conjugate axis is 4 and which passes through the point (1, – 2). Find also the equation of the conjugate hyperbola.
Answer: Let the equation of the hyperbola be `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)` (assuming transverse axis along x-axis).
The problem statement in the solution (not the question) indicates the conjugate axis length is 5. We will follow the solution steps.
Given length of conjugate axis `\(2b = 5 \implies b = \frac{5}{2}\)`. So, `\(b^2 = \frac{25}{4}\)`.
Substitute `\(b^2\)` into the equation:
`\( \frac{x^2}{a^2} - \frac{y^2}{25/4} = 1 \implies \frac{x^2}{a^2} - \frac{4y^2}{25} = 1 \)`
The hyperbola passes through the point `\((1, -2)\)`. Substitute `\(x=1\)` and `\(y=-2\)` into the equation:
`\( \frac{1^2}{a^2} - \frac{4(-2)^2}{25} = 1 \)`
`\( \frac{1}{a^2} - \frac{4 \times 4}{25} = 1 \)`
`\( \frac{1}{a^2} - \frac{16}{25} = 1 \)`
`\( \frac{1}{a^2} = 1 + \frac{16}{25} = \frac{25 + 16}{25} = \frac{41}{25} \)`
So, `\( a^2 = \frac{25}{41} \)`.
Now, substitute `\(a^2\)` and `\(b^2\)` back into the standard equation to get the required hyperbola:
`\( \frac{x^2}{25/41} - \frac{y^2}{25/4} = 1 \)`
`\( \frac{41x^2}{25} - \frac{4y^2}{25} = 1 \)`
Multiply the entire equation by 25:
`\( 41x^2 - 4y^2 = 25 \)`
This is the required equation of the hyperbola.

The **equation of the conjugate hyperbola** to `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)` is `\( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)` (or `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \)`).
Using the `\(a^2\)` and `\(b^2\)` values we found:
`\( \frac{y^2}{25/4} - \frac{x^2}{25/41} = 1 \)`
`\( \frac{4y^2}{25} - \frac{41x^2}{25} = 1 \)`
Multiply the entire equation by 25:
`\( 4y^2 - 41x^2 = 25 \)`
This is the equation of the conjugate hyperbola. The conjugate hyperbola always has the same `\(a\)` and `\(b\)` values, but the roles of \(x\) and \(y\) are swapped in the equation.
In simple words: We used the given conjugate axis length and the point the hyperbola passes through to find the values for \(a^2\) and \(b^2\) for its standard equation. Then we wrote down the hyperbola's equation. After that, we found the equation of its "conjugate" hyperbola, which is related to the first one by swapping the positive and negative terms.

🎯 Exam Tip: Be careful when identifying the length of the conjugate axis; it is `\(2b\)` for a hyperbola with a horizontal transverse axis. Also, remember that the equation of the conjugate hyperbola is found by interchanging the `\(x^2\)` and `\(y^2\)` terms and their denominators, or simply changing the sign of the constant on the right side from 1 to -1, then adjusting signs.

 

Question 12. Find the locus of a point such that the difference of its distances from (4,0) and (- 4, 0) is always equal to 2. Name the curve.
Answer: Let \(P(x, y)\) be the point. Let the two given points be \(F_1(4, 0)\) and \(F_2(-4, 0)\).
The problem states that the difference of the distances from \(P\) to \(F_1\) and \(F_2\) is always equal to 2.
This is the definition of a hyperbola: `\(|PF_1 - PF_2| = 2a\)` (where `\(2a\)` is the length of the transverse axis).
Given `\(|PF_1 - PF_2| = 2\)`, so `\(2a = 2 \implies a = 1\)`. Thus, `\(a^2 = 1\)`.
The foci of the hyperbola are `\(F_1(4, 0)\)` and `\(F_2(-4, 0)\)`. For a hyperbola centered at the origin with transverse axis on the x-axis, the foci are `\((\pm ae, 0)\)`.
So, `\(ae = 4\).
Substitute `\(a = 1\)` into `\(ae = 4\)`:
`\( 1 \times e = 4 \implies e = 4 \)`
Since `\(e = 4 > 1\)`, the curve is indeed a hyperbola.

Now, we find `\(b^2\)` using the relationship `\(b^2 = a^2(e^2 - 1)\)`:
`\( b^2 = 1^2 (4^2 - 1) \)`
`\( b^2 = 1 (16 - 1) = 15 \)`
The equation of a hyperbola with transverse axis along the x-axis and center at the origin is `\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)`
Substitute `\(a^2 = 1\)` and `\(b^2 = 15\)`:
`\( \frac{x^2}{1} - \frac{y^2}{15} = 1 \)`
`\( 15x^2 - y^2 = 15 \)`
This is the required equation of the locus.
The name of the curve is a **Hyperbola**. This problem beautifully illustrates the geometric definition of a hyperbola, where the difference in distances to two fixed points remains constant.
In simple words: We know that any point on a hyperbola has a constant difference in distance from two fixed points called foci. Since the problem gave us these two fixed points and the constant difference, we could find the values of \(a\) and \(e\). From these, we calculated \(b^2\) and then wrote the standard equation for the hyperbola. The curve is a hyperbola.

🎯 Exam Tip: Recognize the definition `\(|PF_1 - PF_2| = 2a\)` as fundamental for a hyperbola. This immediately gives you `\(a\)` and helps identify the foci. Always verify that `\(e > 1\)` for the curve to be a hyperbola.

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