Q.10. Let an ellipse , , pass through the point and have eccentricity . Then the length of its latu

Q: What is the correct answer for JEE Main Mathematics Q10?

The correct answer is D. Sol. The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a < b$. This means the major axis is along the y-axis. For such an ellipse, the eccentricity $e$ is given by $e^2 = 1 - \frac{a^2}{b^2}$. Given $e = \frac{\sqrt{5}}{3}$, so $e^2 = \frac{5}{9}$. $\frac{5}{9} = 1 - \frac{a^2}{b^2} \Rightarrow \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$. So, $a^2 = \frac{4}{9}b^2$.

Q: Q.10. Let an ellipse , , pass through the point and have eccentricity . Then the length of its latus rectum is :

The correct answer is D. Sol. The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a < b$. This means the major axis is along the y-axis. For such an ellipse, the eccentricity $e$ is given by $e^2 = 1 - \frac{a^2}{b^2}$. Given $e = \frac{\sqrt{5}}{3}$, so $e^2 = \frac{5}{9}$. $\frac{5}{9} = 1 - \frac{a^2}{b^2} \Rightarrow \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$. So, $a^2 = \frac{4}{9}b^2$.

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

A. 4\sqrt{5}/3 · B. 2\sqrt{5} · C. 7\sqrt{5}/3 · D. 8\sqrt{5}/3

Question

Q.10. Let an ellipse , , pass through the point  and have eccentricity . Then the length of its latus rectum is :

FirstInTest · Accepted Answer

Sol. The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a < b$. This means the major axis is along the y-axis.
For such an ellipse, the eccentricity $e$ is given by $e^2 = 1 - \frac{a^2}{b^2}$.
Given $e = \frac{\sqrt{5}}{3}$, so $e^2 = \frac{5}{9}$.
$\frac{5}{9} = 1 - \frac{a^2}{b^2} \Rightarrow \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$.
So, $a^2 = \frac{4}{9}b^2$.

The ellipse passes through the point $(4, 3)$. Substitute these coordinates into the ellipse equation:
$\frac{4^2}{a^2} + \frac{3^2}{b^2} = 1$
$\frac{16}{a^2} + \frac{9}{b^2} = 1$
Substitute $a^2 = \frac{4}{9}b^2$ into this equation:
$\frac{16}{\frac{4}{9}b^2} + \frac{9}{b^2} = 1$
$\frac{16 	imes 9}{4b^2} + \frac{9}{b^2} = 1$
$\frac{36}{b^2} + \frac{9}{b^2} = 1$
$\frac{45}{b^2} = 1 \Rightarrow b^2 = 45$.

Now find $a^2$:
$a^2 = \frac{4}{9}b^2 = \frac{4}{9}(45) = 4 	imes 5 = 20$.

So, the equation of the ellipse is $\frac{x^2}{20} + \frac{y^2}{45} = 1$.

For an ellipse with major axis along the y-axis ($a < b$), the length of the latus rectum is $\frac{2a^2}{b}$.
Length of latus rectum $= \frac{2(20)}{\sqrt{45}} = \frac{40}{\sqrt{9 	imes 5}} = \frac{40}{3\sqrt{5}}$.
Rationalize the denominator:
$= \frac{40}{3\sqrt{5}} 	imes \frac{\sqrt{5}}{\sqrt{5}} = \frac{40\sqrt{5}}{3 	imes 5} = \frac{8\sqrt{5}}{3}$.

Answer

4\sqrt{5}/3

Answer

2\sqrt{5}

Answer

7\sqrt{5}/3

Answer

8\sqrt{5}/3

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