Let $P(10,\,2\sqrt{15})$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, whose foci are $S$ and $S’$. If the length of its latus rectum is $8$, then the square of the area of $\triangle PSS’$ is equal to:
A) 4200
B) 1462
C) 900
D) 2700 ✓
✓
Solution Steps
1
Use the latus rectum condition
Latus rectum of $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is $\dfrac{2b^2}{a} = 8$
$$b^2 = 4a \quad \cdots(1)$$
2
P(10, 2√15) lies on the hyperbola
$$\frac{100}{a^2} – \frac{(2\sqrt{15})^2}{b^2} = 1$$
$$\frac{100}{a^2} – \frac{60}{b^2} = 1 \quad \cdots(2)$$
3
Solve for a
Substitute $b^2 = 4a$ into (2):
$$\frac{100}{a^2} – \frac{60}{4a} = 1 \implies \frac{100}{a^2} – \frac{15}{a} = 1$$
Let $u = \dfrac{1}{a}$:
$$100u^2 – 15u – 1 = 0$$
$$(20u+1)(5u-1)=0 \implies u = \frac{1}{5} \;\;(\text{taking positive})$$
$a = 5,\quad b^2 = 4a = 20$
4
Find the foci
$$c^2 = a^2 + b^2 = 25 + 20 = 45 \implies c = 3\sqrt{5}$$
Foci: $S = (3\sqrt{5},\,0)$ and $S’ = (-3\sqrt{5},\,0)$
$$SS’ = 2c = 6\sqrt{5}$$
5
Area of △PSS’
$S$ and $S’$ both lie on the $x$-axis, so base $= SS’ = 6\sqrt{5}$
Height $=$ perpendicular distance from $P$ to $x$-axis $= |y_P| = 2\sqrt{15}$
$$\text{Area} = \frac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15}$$
$$= 6\sqrt{5\times 15} = 6\sqrt{75} = 6 \times 5\sqrt{3} = 30\sqrt{3}$$
6
Square of the area
$$(\text{Area})^2 = (30\sqrt{3})^2 = 900 \times 3 = \boxed{2700}$$
✓ Answer: D) 2700
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Key Insight
Latus rectum=8 → b²=4a. Point on hyperbola → a=5, b²=20, c=3√5. Area=(½)×6√5×2√15=30√3 → Area²=2700.
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Theory: Hyperbola — All Key Formulas
Standard Hyperbola: x²/a² − y²/b² = 1
| Property | Formula |
|---|---|
| Foci | $(\pm c,\,0)$ where $c^2=a^2+b^2$ |
| Eccentricity | $e = c/a > 1$ |
| Latus Rectum | $\ell = \dfrac{2b^2}{a}$ |
| Directrices | $x = \pm\dfrac{a}{e}$ |
| Focal distances | $r_1 = |ex-a|$, $r_2 = |ex+a|$ |
| Focal property | $|PS – PS’| = 2a$ |
| Transverse axis | Length $= 2a$ (along $x$-axis) |
| Conjugate axis | Length $= 2b$ (along $y$-axis) |
Latus Rectum — Derivation
The latus rectum is the chord through a focus perpendicular to the transverse axis.
At $x = c$: $\dfrac{c^2}{a^2} – \dfrac{y^2}{b^2} = 1 \Rightarrow y^2 = b^2\!\left(\dfrac{c^2}{a^2}-1\right) = \dfrac{b^4}{a^2}$
So $y = \pm\dfrac{b^2}{a}$. Length of latus rectum $= \dfrac{2b^2}{a}$.
At $x = c$: $\dfrac{c^2}{a^2} – \dfrac{y^2}{b^2} = 1 \Rightarrow y^2 = b^2\!\left(\dfrac{c^2}{a^2}-1\right) = \dfrac{b^4}{a^2}$
So $y = \pm\dfrac{b^2}{a}$. Length of latus rectum $= \dfrac{2b^2}{a}$.
Area of Triangle with Focal Chord
When both foci $S(\!+c,0)$ and $S'(-c,0)$ are on the $x$-axis and $P=(x_0,y_0)$:
$$\text{Area}(\triangle PSS’) = \frac{1}{2}|SS’|\cdot|y_0| = \frac{1}{2}\cdot 2c\cdot|y_0| = c\cdot|y_0|$$
Here: $c = 3\sqrt{5}$, $y_0 = 2\sqrt{15}$
Area $= 3\sqrt{5}\times 2\sqrt{15} = 6\sqrt{75} = 30\sqrt{3}$
Area² $= 900\times 3 = 2700$
$$\text{Area}(\triangle PSS’) = \frac{1}{2}|SS’|\cdot|y_0| = \frac{1}{2}\cdot 2c\cdot|y_0| = c\cdot|y_0|$$
Here: $c = 3\sqrt{5}$, $y_0 = 2\sqrt{15}$
Area $= 3\sqrt{5}\times 2\sqrt{15} = 6\sqrt{75} = 30\sqrt{3}$
Area² $= 900\times 3 = 2700$
Hyperbola vs Ellipse: Key Differences
| Property | Hyperbola | Ellipse |
|---|---|---|
| Equation | $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ | $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ |
| $c^2$ | $a^2+b^2$ | $a^2-b^2$ |
| $e$ | $>1$ | $<1$ |
| Focal property | $|PS-PS’|=2a$ | $PS+PS’=2a$ |
| Latus rectum | $2b^2/a$ | $2b^2/a$ |
Parametric Form & Useful Identities
Parametric: $x = a\sec\theta$, $y = b\tan\theta$
Focal radii: For point $(x_1,y_1)$ on right branch:
$PS = |ex_1 – a|$, $\;PS’ = |ex_1 + a|$
Verification here: $e = c/a = 3\sqrt{5}/5$
$PS’ – PS = e\cdot x_1 + a – (e\cdot x_1 – a) = 2a = 10$ ✓
Product: $PS\cdot PS’ = \sqrt{(PS’^2)(PS^2)} = \sqrt{(205+60\sqrt{5})(205-60\sqrt{5})}$
$= \sqrt{205^2 – (60\sqrt{5})^2} = \sqrt{42025 – 18000} = \sqrt{24025} = 155$
Focal radii: For point $(x_1,y_1)$ on right branch:
$PS = |ex_1 – a|$, $\;PS’ = |ex_1 + a|$
Verification here: $e = c/a = 3\sqrt{5}/5$
$PS’ – PS = e\cdot x_1 + a – (e\cdot x_1 – a) = 2a = 10$ ✓
Product: $PS\cdot PS’ = \sqrt{(PS’^2)(PS^2)} = \sqrt{(205+60\sqrt{5})(205-60\sqrt{5})}$
$= \sqrt{205^2 – (60\sqrt{5})^2} = \sqrt{42025 – 18000} = \sqrt{24025} = 155$
❓
Frequently Asked Questions
Q
Why is c²=a²+b² for hyperbola (not a²−b²)?
▼
For ellipse, c²=a²−b² because foci are inside (c<a). For hyperbola, foci are outside (c>a), so c²=a²+b². Remember: hyperbola has a “+” sign between a² and b² for c², ellipse has “−”.
Q
How do we factor 100u²−15u−1=0?
▼
Product = 100×(−1) = −100. We need two numbers that multiply to −100 and add to −15. Those are −20 and +5. So: 100u²−20u+5u−1 = 20u(5u−1)+1(5u−1) = (20u+1)(5u−1). Roots: u=−1/20 (rejected, a>0) and u=1/5, so a=5.
Q
Why do we reject u = −1/20?
▼
u = 1/a. Since a is a positive length parameter of the hyperbola, u must be positive. u = −1/20 gives a = −20, which is not physically valid.
Q
How to simplify 6√75?
▼
√75 = √(25×3) = 5√3. So 6√75 = 6×5√3 = 30√3. Squaring: (30√3)² = 900×3 = 2700.
Q
Can we verify using the focal distance formula?
▼
e = c/a = 3√5/5. PS’ = ex₀+a = (3√5/5)×10+5 = 6√5+5. PS = ex₀−a = 6√5−5. Then PS’−PS=10=2a ✓. Area via ½|PS||PS’|sin∠SPS’ also gives 2700.
Q
What is the eccentricity of this hyperbola?
▼
e = c/a = 3√5/5 = (3/5)√5 ≈ 1.342. Since e>1, it confirms this is a hyperbola. The eccentricity tells us how “open” the branches are.
Q
What is the asymptote equation of this hyperbola?
▼
Asymptotes of x²/a²−y²/b²=1 are y=±(b/a)x. Here: y=±(√20/5)x = ±(2√5/5)x = ±(2/√5)x. The hyperbola approaches these lines at infinity.
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