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Hyperbola question

2016 · 9 Apr · Shift 1 · Q33

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Hyperbola
2016 · 9 Apr · Shift 1 · Q33

Hyperbola question

2016 · 9 Apr · Shift 1 · Q33

JEE MainMathematicsHyperbolaMCQ+4 / −1

Let a and b respectively be the semitransverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9e2 − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a2 − b2 is equal to :

A
7
B
$-$ 7
C
5
D
$-$ 5

View written solutionFree

Correct answer: B

Find the eccentricity $e$

Given $9e^2-18e+5=0$

Solve:

=\frac{18\pm\sqrt{324-180}}{18} =\frac{18\pm 12}{18}$$ So, $$e=\frac{30}{18}=\frac53 \quad \text{or} \quad e=\frac{6}{18}=\frac13$$ For a hyperbola, eccentricity satisfies $e>1$, hence $$e=\frac53$$ --- 2. **Use the directrix formula** Since the focus is $S(5,0)$, the hyperbola is of the form $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ with transverse axis along the $x$-axis. For this hyperbola: - foci are $(\pm ae,0)$ - directrices are $x=\pm \frac{a}{e}$ Given the corresponding directrix is $$5x=9 \implies x=\frac95$$ Also the focus is $(5,0)$, so $$ae=5$$ Since $e=\frac53$, $$a\cdot \frac53=5 \implies a=3$$ Check with directrix: $$\frac{a}{e}=\frac{3}{5/3}=\frac95$$ which matches. --- 3. **Find $b^2$ using hyperbola relation** For a hyperbola, $$e^2=1+\frac{b^2}{a^2}$$ So, $$b^2=a^2(e^2-1)$$ Now, $$a^2=9, \qquad e^2=\left(\frac53\right)^2=\frac{25}{9}$$ Thus, $$b^2=9\left(\frac{25}{9}-1\right)=9\left(\frac{16}{9}\right)=16$$ --- 4. **Compute $a^2-b^2$** $$a^2-b^2=9-16=-7$$ --- 5. **Option check** - A: $7$ ❌ - B: $-7$ ✅ - C: $5$ ❌ - D: $-5$ ❌ Therefore, the correct answer is $$\boxed{-7}$$

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