Question

# The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is

## The correct answer is:

### Related Questions to study

### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

Finding the smallest common multiple between any two or more numbers is done using the least common multiple (LCM) approach. A number that is a multiple of two or more other numbers is said to be a common multiple. Here we understood the concept of LCM and the pairs, so the total pairs can be 225.

### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

Finding the smallest common multiple between any two or more numbers is done using the least common multiple (LCM) approach. A number that is a multiple of two or more other numbers is said to be a common multiple. Here we understood the concept of LCM and the pairs, so the total pairs can be 225.

### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Here we used the concept of number system and the rectangle, we can also solve it by permutation and combination. herefore, we get the number of rectangles possible with odd side length = m^{2}n^{2}.

### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Here we used the concept of number system and the rectangle, we can also solve it by permutation and combination. herefore, we get the number of rectangles possible with odd side length = m^{2}n^{2}.

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

The coefficient of in is

The coefficient of in is