
To solve the problem, we need to find the ratio of volumes of a right circular cylinder and a cone given the ratios of their radii and heights.
1. Volume Formulas:
The volume of a cylinder is given by:
$ V_{\text{cylinder}} = \pi r_1^2 h_1 $
The volume of a cone is given by:
$ V_{\text{cone}} = \frac{1}{3} \pi r_2^2 h_2 $
2. Given Ratios:
Ratio of radii: $r_1 : r_2 = 2 : 3$
Ratio of heights: $h_1 : h_2 = 3 : 4$
3. Expressing Volumes in Terms of Ratios:
Let $r_1 = 2$, $r_2 = 3$, $h_1 = 3$, $h_2 = 4$
Then:
$V_{\text{cylinder}} = \pi (2)^2 (3) = \pi \cdot 4 \cdot 3 = 12\pi$
$V_{\text{cone}} = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi \cdot 9 \cdot 4 = 12\pi$
4. Finding the Ratio:
$ \frac{V_{\text{cylinder}}}{V_{\text{cone}}} = \frac{12\pi}{12\pi} = 1 : 1 $
Final Answer:
The ratio of their volumes is $1 : 1$
If \( S \) and \( S' \) are the foci of the ellipse \[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \] and \( P \) is a point on the ellipse, then \[ \min (SP \cdot S'P) + \max (SP \cdot S'P) \] is equal to:
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to: