Show that the height of the cylinder of maximum volume, that can be inscribed in a sphere of radius R is Also find the maximum volume.

OR

Find the equation of the normal at a point on the curve x² = 4y which passes through the point (1, 2). Also find the equation of the corresponding tangent.

Given, radius of the sphere is R.

Let r and h be the radius and the height of the inscribed cylinder respectively.

Let Volume of cylinder = V

and we know that,

Volume of cylinder = πr²h

Now, we find the h

In ∆ABC,

Using Pythagoras theorem

(AC)² = (AB)² + (BC)²

⇒ (R + R)² = (r + r)² + h²

⇒ (2R)² = (2r)² + h²

⇒ 4R² – 4r² = h²

⇒ h² = 4(R² – r²)

⇒ h = 2√(R² – r²)

So,

V = πr²h

= πr²[2√(R² – r²)]

= 2πr²√(R² – r²)

Differentiating the above with respect to r, we get

For maxima or minima,

⇒ 4πR²r – 6πr³ = 0

⇒ 4πR²r = 6πr³

⇒ 2R² = 3r²

Now, we again differentiate with respect to x, we get

[here we apply the quotient rule]

Now, when

∴Volume is the maximum when

When , then h = 2√(R² – r²)

Hence, the volume of the cylinder is the maximum when the height of the cylinder is

OR

The equation of the given curve is x² = 4y.

Differentiating with respect to x, we get

Let (h, k) be the coordinates of the point of contact of the normal to the curve x² = 4y.

Now, slope of the tangent at (h, k) is given by

Hence, slope of the normal at (h, k)

We know that,

Slope of tangent × Slope of normal = -1

We know that,

Equation of line at (x₁, y₁) and having slope m is

(y – y₁) = m(x – x₁)

Therefore, the equation of normal at (h, k) is

(y – y₁) = m(x – x₁)

…(i)

Since, it passes through the point (1, 2) we have

…(ii)

Now, (h, k) lies on the curve x² = 4y

So, we have:

h² = 4k …(iii)

Solving (ii) and (iii), we get,

⇒ h³ = 8

⇒ h = 2

Substitute the value of h = 2 in eq. (iii), we get

(2)² = 4k

⇒ k = 1

From (i), the required equation of normal is:

⇒ y – 1 = -(x – 2)

⇒ y – 1 = – x + 2

⇒ x + y = 3

Also, slope of the tangent

∴ Equation of tangent at (1, 2) and having slope 1, we have

y – 2 = (1)(x – 1)

⇒ y – 2 = x – 1

⇒ y = x + 1