Ncert Solutions Maths class 11th
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a month agoContributor-Level 10
Given the family of parabolas y² = 4a (x+a).
Differentiate with respect to x:
2y (dy/dx) = 4a
a = (y/2) (dy/dx)
Substitute a back into the original equation:
y² = 4 * (y/2) (dy/dx) * [x + (y/2) (dy/dx)]
y² = 2y (dy/dx) * [x + (y/2) (dy/dx)]
y = 2 (dy/dx) * [x + (y/2) (dy/dx)]
y = 2x (dy/dx) + y (dy/dx)²
y (dy/dx)² + 2x (dy/dx) - y = 0
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a month agoNew question posted
a month agoNew question posted
a month agoNew answer posted
a month agoContributor-Level 10
The general equation of a circle is given by:
az z? + α? z + αz? + d = 0
This can be rewritten as:
z? + (α? /a)z + (α/a)z? + d/a = 0
From this, we can identify the centre and radius:
Centre = -α/a
Radius = √ (|-α/a|² - d/a)
For a real circle to exist, the term under the square root must be non-negative:
|-α/a|² - d/a ≥ 0
|α|²/|a|² - d/a ≥ 0
|α|² - ad ≥ 0, where a ∈ R - {0}.
New answer posted
a month agoContributor-Level 10
We need to find the remainder of (2021)³? ² when divided by 17.
First, find the remainder of 2021 divided by 17.
2021 = 17 * 118 + 15.
So, 2021 ≡ 15 (mod 17).
Also, 15 ≡ -2 (mod 17).
So, (2021)³? ² ≡ (-2)³? ² (mod 17).
(-2)³? ² = 2³? ² = (2? )? ⋅ 2² = 16? ⋅ 4.
Since 16 ≡ -1 (mod 17),
16? ⋅ 4 ≡ (-1)? ⋅ 4 (mod 17).
≡ 1 ⋅ 4 (mod 17)
≡ 4 (mod 17).
The remainder is 4.
New answer posted
a month agoContributor-Level 10
Given equation of tangent is 2x - y + 1 = 0
equation of normal is x + 2y = 12
Solving with x - 2y = 4 we get centre at (6,2) radius = √ (36 + 9) = √45 = 3√5.
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