Q.11. If the eccentricity of the hyperbola , passing through , satisfies , then the length of the l

Q: What is the correct answer for JEE Main Mathematics Q11?

The correct answer is A. Sol. Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ It passes through $(6,4\sqrt{3})$ $\frac{36}{a^2} - \frac{48}{b^2} = 1$ ...(1) Also $15(e^2 + 1) = 34e$ $15e^2 - 34e + 15 = 0$ $(5e-3)(3e-5) = 0$ $e = \frac{3}{5}$ or $e = \frac{5}{3}$ For hyperbola $e > 1$, so $e = \frac{5}{3}$ We know $e^2 = 1 + \frac{b^2}{a^2} \Rightarrow \frac{25}{9} = 1 + \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = \fr

Q: Q.11. If the eccentricity of the hyperbola , passing through , satisfies , then the length of the latus rectum of the hyperbola is:

The correct answer is A. Sol. Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ It passes through $(6,4\sqrt{3})$ $\frac{36}{a^2} - \frac{48}{b^2} = 1$ ...(1) Also $15(e^2 + 1) = 34e$ $15e^2 - 34e + 15 = 0$ $(5e-3)(3e-5) = 0$ $e = \frac{3}{5}$ or $e = \frac{5}{3}$ For hyperbola $e > 1$, so $e = \frac{5}{3}$ We know $e^2 = 1 + \frac{b^2}{a^2} \Rightarrow \frac{25}{9} = 1 + \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = \fr

Q: What are the options for this JEE Main Mathematics question?

A. 10 · B. 20 · C. 25 · D. 30

Question

Q.11. If the eccentricity  of the hyperbola , passing through , satisfies , then the length of the latus rectum of the hyperbola  is:

FirstInTest · Accepted Answer

Sol. Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
It passes through $(6,4\sqrt{3})$
$\frac{36}{a^2} - \frac{48}{b^2} = 1$ ...(1)
Also $15(e^2 + 1) = 34e$
$15e^2 - 34e + 15 = 0$
$(5e-3)(3e-5) = 0$
$e = \frac{3}{5}$ or $e = \frac{5}{3}$
For hyperbola $e > 1$, so $e = \frac{5}{3}$
We know $e^2 = 1 + \frac{b^2}{a^2} \Rightarrow \frac{25}{9} = 1 + \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = \frac{16}{9}$ ...(2)
From (1) and (2):
$\frac{36}{a^2} - \frac{48}{b^2} = 1 \Rightarrow \frac{36}{a^2} - \frac{48}{\frac{16}{9}a^2} = 1 \Rightarrow \frac{36}{a^2} - \frac{48 \cdot 9}{16a^2} = 1$
$\Rightarrow \frac{36}{a^2} - \frac{3 \cdot 9}{a^2} = 1 \Rightarrow \frac{36-27}{a^2} = 1 \Rightarrow \frac{9}{a^2} = 1 \Rightarrow a^2 = 9$
From (2), $b^2 = \frac{16}{9}a^2 = \frac{16}{9} \cdot 9 = 16$
Now consider the second hyperbola: $\frac{x^2}{b^2} - \frac{y^2}{2(a^2+1)} = 1$
Here, $A^2 = b^2 = 16$ and $B^2 = 2(a^2+1) = 2(9+1) = 20$
Length of latus rectum $= \frac{2B^2}{A} = \frac{2 \cdot 20}{\sqrt{16}} = \frac{40}{4} = 10$

Answer

10

Answer

20

Answer

25

Answer

30

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions