Class 12 RD Sharma Solutions - Chapter 16 Tangents and Normals - Exercise 16.3

Exercise 16.3 in Chapter 16 of RD Sharma's Class 12 mathematics textbook delves into more advanced applications of tangents and normals. This section covers problems like finding the length of tangents, determining the points of contact for tangents drawn from external points, and solving problems involving the properties of normals to various curves.

Students are expected to apply their knowledge of calculus, coordinate geometry, and algebraic manipulation to solve these complex problems.

Question 1. Find the angle of intersection of the following curves:

(i) y² = x and x² = y

Solution:

First curve is y² = x. . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 1

=> m₁ = dy/dx = 1/2y

Second curve is x² = y . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x = dy/dx

=> m₂ = dy/dx = 2x

Solving (1) and (2), we get,

=> x⁴ − x = 0

=> x (x³ − 1) = 0

=> x = 0 or x = 1

We know that the angle of intersection of two curves is given by,

tan θ =|\frac{m_1-m_2}{1+m_1m_2}|

where m₁ and m₂ are the slopes of the curves.

When x = 0, then y = 0.

So, m₁ = 1/2y = 1/0 = ∞

m₂ = 2x = 2(0) = 0

Therefore, tan θ =|\frac{(∞−0)}{1+ 0×∞}| = ∞

=> θ = π/2

When x = 1, then y = 1.

So, m₁ = 1/2y = 1/2

m₂ = 2x = 2(1) = 2

Therefore, tan θ =\frac{2-\frac{1}{2}}{1+2×\frac{1}{2}} = \frac{3}{4}

=> θ = tan⁻¹ (3/4)

(ii) y = x² and x² + y² = 20

Solution:

First curve is y = x². . . . . (1)

Differentiating both sides with respect to x, we get,

=> (dy/dx) = 2x

=> m₁ = dy/dx = 2x

Second curve is x² + y² = 20 . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) = 0

=> m₂ = dy/dx = −x/y

Solving (1) and (2), we get,

=> y² +y − 20 = 0

=> y² + 5y − 4y − 20 = 0

=> (y + 5) (y − 4) = 0

=> y = −5 or y = 4

Ignoring y = − 5 as x becomes √(−5) in that case, which is not possible.

When y = 4, we get x² = 4

=> x = ±2

We know that the angle of intersection of two curves is given by,

tan θ =|\frac{m_1-m_2}{1+m_1m_2}|

where m₁ and m₂ are the slopes of the curves.

When x = ±2 and y = 4, we get,

m₁ = 2x = 2(2) = 4 or ±4

m₂ = −x/y = −2/4 = −1/2

So, tan θ =|\frac{4+\frac{1}{2}}{1+4×(-\frac{1}{2})}| = \frac{9}{2}

=> θ = tan⁻¹ (9/2)

When x = −2 and y = 4, we get,

m₁ = 2x = 4 or −4

m₂ = −x/y = 1/2 or −1/2

So, tan θ =|\frac{-4-\frac{1}{2}}{1-4×(\frac{1}{2})}| = \frac{9}{2}

=> θ = tan⁻¹ (9/2)

(iii) 2y² = x³ and y² = 32x

Solution:

First curve is 2y² = x³. . . . . (1)

Differentiating both sides with respect to x, we get,

=> 4y (dy/dx) = 3x²

=> m₁ = dy/dx = 3x²/4y

Second curve is y² = 32x. . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 32

=> m₂ = dy/dx = 32/2y = 16/y

Solving (1) and (2), we get,

=> 2(32x) = x³

=> x³ − 64x = 0

=> x(x² − 64) = 0

=> x = 0 or x² − 64 = 0

=> x = 0 or x = ±8

We know that the angle of intersection of two curves is given by,

tan θ =|\frac{m_1-m_2}{1+m_1m_2}|

where m₁ and m₂ are the slopes of the curves.

When x = 0 then y = 0.

m₁ = 3x²/4y = ∞

m₂ = 16/y = ∞

So, tan θ = ∞

=> θ = π/2

When x = ±8, then y = ±16.

m₁ = 3x²/4y = 3 or −3

m₂ = 16/y = 1 or −1

So, tan θ =|\frac{3-1}{1+3×1}| = \frac{1}{2}

=> θ = tan⁻¹ (1/2)

(iv) x² + y² – 4x – 1 = 0 and x² + y² – 2y – 9 = 0

Solution:

First curve is x² + y² – 4x – 1 = 0. . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) – 4 = 0

=> m₁ = dy/dx = (2–x)/y

Second curve is x² + y² – 2y – 9 = 0. . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) – 2 (dy/dx) = 0

=> m₂ = dy/dx = –x/(y–1)

First curve can be written as,

=> (x – 2)² + y² – 5 = 0 . . . . (3)

Subtracting (2) from (1), we get

=> x² + y² – 4x – 1 – x² – y² + 2y + 9 = 0

=> – 4x – 1 + 2y + 9 = 0

=> 2y = 4x – 8

=> y = 2x – 4

Putting y = 2x – 4 in (1), we get,

=> (x – 2)² + (2x – 4)² – 5 = 0

⇒ (x – 2)²(1 + 4) – 5 = 0

⇒ 5(x – 2)² – 5 = 0

⇒ (x – 2)² = 1

⇒ x = 3 or x = 1

So, when x = 3 then y = 6 – 4 = 2

m₁ = (2–x)/y = (2–3)/2 = –1/2

m₂ = –x/(y–1) = –3/(2–1) = –3

So, tan θ =|\frac{\frac{-1}{2}+3}{1+\frac{-1}{2}×(-3)}| = 1

=> θ = π/4

So, when x = 1 then y = 2 – 4 = – 2

m₁ = (2–x)/y = (2–1)/(–2) = –1/2

m₂ = –x/(y–1) = –1/(–2–1) = 1/3

So, tan θ =|\frac{\frac{-1}{2}-\frac{1}{3}}{1+\frac{-1}{2}×\frac{1}{3}}| = 1

=> θ = π/4

(v) x²/a² + y²/b² = 1 and x² + y² = ab

Solution:

First curve is x²/a² + y²/b² = 1 . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x/a² + (2y/b²) (dy/dx) = 0

=> m₁ = dy/dx = –b²x/a²y

Second curve is x² + y² = ab . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) = 0

=> m₂ = dy/dx = –2x/2y = –x/y

Solving (1) and (2), we get,

=> x²/a² + (ab – x²)/b² = 1

=> x²b² – a²x² = a²b² – a³b

=> x² =\frac{a^2b(b-a)}{b^2-a^2}

=> x =\pm\sqrt{\frac{a^2b}{a+b}}

From (2), we get, y² =ab-\frac{a^2b}{a+b}

=> y =\pm\sqrt{\frac{ab^2}{a+b}}

So, m₁ = –b²x/a²y =\frac{-b^2\sqrt{\frac{a^2b}{a+b}}}{a^2\sqrt{\frac{ab^2}{a+b}}}

=\frac{-b\sqrt{b}}{a\sqrt{a}}

m₂ = –x/y =\frac{-\sqrt{\frac{a^2b}{a+b}}}{\sqrt{\frac{ab^2}{a+b}}}

=\frac{-\sqrt{a}}{\sqrt{b}}

Therefore, tan θ =\frac{|\frac{-b\sqrt{b}}{a\sqrt{a}}+\sqrt{\frac{a}{b}}|}{|1+\frac{a}{b}|}

=> tan θ =\frac{|\frac{a^2-b^2}{a\sqrt{a}b}|}{|\frac{a+b}{b}|}

=> tan θ =\frac{a-b}{\sqrt{a}b}

=> θ = tan^–1 ((a–b)/√ab)

(vi) x² + 4y² = 8 and x² – 2y² = 2

Solution:

First curve is x² + 4y² = 8 . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x + 8y (dy/dx) = 0

=> m₁ = dy/dx = –2x/8y = –x/4y

Second curve is x² – 2y² = 2 . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x – 4y (dy/dx) = 0

=> m₂ = dy/dx = x/2y

Solving (1) and (2), we get,

6y² = 6 => y2 = ±1

x² = 2 + 2 => x = ±2

So, tan θ =|\frac{1-2}{1+1×2}|=\frac{1}{3}

=> θ = tan^–1 (1/3)

(vii) x² = 27y and y² = 8x

Solution:

First curve is x² = 27y . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x = 27 (dy/dx)

=> m₁ = dy/dx = 2x/27

Second curve is y² = 8x . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 8

=> m₂ = dy/dx = 8/2y = 4/y

Solving (1) and (2), we get,

=> y⁴/64 = 27y

=> y (y³ − 1728) = 0

=> y = 0 or y = 12

And x = 0 or x = 18.

So, when x = 0 and y = 0

m₁ = 0 and m₂ = ∞

tan θ =|\frac{0-∞}{1+0×∞}| = ∞

=> θ = π/2

So, when x = 18 and y = 12

m₁ = 2x/27 = 12/9 = 4/3 and m₂ = 4/y = 1/3

tan θ =|\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{3}×\frac{1}{3}}|=\frac{9}{13}

=> θ = tan⁻¹ (9/13)

(viii) x² + y² = 2x and y² = x

Solution:

First curve is x² + y² = 2x . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) = 2

=> m₁ = dy/dx = (1–x)/y

Second curve is y² = x . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 1

=> m₂ = dy/dx = 1/2y

Solving (1) and (2), we get,

=> x² – x = 0

=> x = 0 or x = 1

And y = 0 or y = ±1.

When x = 0, y = 0, m₁ = ∞ and m₂ = ∞

tan θ =|\frac{∞-∞}{1+∞×∞}| = ∞

=> θ = π/2

When x = 1 and y = ±1, m₁ = 0 and m₂ = 1/2

tan θ =|\frac{0-\frac{1}{2}}{1+0×\frac{1}{2}}|=\frac{1}{2}

=> θ = tan⁻¹ (1/2)

(ix) y = 4 − x² and y = x²

Solution:

First curve is y = 4 − x² . . . . (1)

Differentiating both sides with respect to x, we get,

=> dy/dx = −2x

=> m₁ = dy/dx = −2x

Second curve is y = x² . . . . (2)

Differentiating both sides with respect to x, we get,

=> dy/dx = 2x

=> m₂ = dy/dx = 2x

Solving (1) and (2), we get,

=> 2x² = 4

=> x = ±√2

And y = 2

So, m₁ = −2x = −2√2 and m₂ = 2x = 2√2

tan θ =|\frac{-2\sqrt{2}-2\sqrt{2}}{1+(-2\sqrt{2})×2\sqrt{2}}|=\frac{4\sqrt{2}}{7}

=> θ = tan⁻¹ (4√2/7)

Question 2. Show that the following set of curves intersect orthogonally:

(i) y = x³ and 6y = 7 – x²

Solution:

First curve is y = x³ . . . . (1)

Differentiating both sides with respect to x, we get,

=> dy/dx = 3x²

=> m₁ = dy/dx = 3x²

Second curve is 6y = 7 – x² . . . . (2)

Differentiating both sides with respect to x, we get,

=> 6y (dy/dx) = – 2x

=> m₂ = dy/dx = –2x/6y = – x/3y

Solving (1) and (2), we get,

=> 6y = 7 – x²

=> 6x³ + x² – 7 = 0

As x = 1 satisfies this equation, we get x = 1 and y = 1³ = 1

So, m₁ = 3 and m₂ = – 1/3

Two curves intersect orthogonally if m₁m₂ = –1

=> 3 × (–1/3) = –1

Hence proved.

(ii) x³ – 3xy² = – 2 and 3x² y – y³ = 2

Solution:

First curve is x³ – 3xy² = – 2

Differentiating both sides with respect to x, we get,

=> 3x² – 3y² – 6xy (dy/dx) = 0

=> m₁ = dy/dx = 3(x²–y²)/6xy

Second curve is 3x²y – y³ = 2

Differentiating both sides with respect to x, we get,

=> 6xy + 3x² (dy/dx) – 3y² (dy/dx) = 0

=> m₂ = dy/dx = –6xy/3(x²–y²)

Two curves intersect orthogonally if m₁m₂ = –1

=>\frac{3(x^2–y^2)}{6xy}×\frac{–6xy}{3(x^2–y^2)} = –1

Hence proved.

(iii) x² + 4y² = 8 and x² – 2y² = 4.

Solution:

First curve is x² + 4y² = 8 . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x + 8y (dy/dx) = 0

=> m₁ = dy/dx = 2x/8y = –x/4y

Second curve is x² – 2y² = 4 . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x – 4y (dy/dx) = 0

=> m₂ = dy/dx = 2x/4y = x/2y

Solving (1) and (2), we get,

=> x = 4/√3 and y = √2/√3

So, m₁ = –x/4y = –1/√2

m₂ = x/2y = √2

Two curves intersect orthogonally if m₁m₂ = –1

=> (–1/√2) × √2 = –1

Hence proved.

Question 3. Show that the curves:

(i) x² = 4y and 4y + x² = 8 intersect orthogonally at (2, 1).

Solution:

First curve is x² = 4y

Differentiating both sides with respect to x, we get,

=> 2x = 4 (dy/dx)

=> m₁ = dy/dx = 2x/4 = x/2

Second curve is 4y + x² = 8

Differentiating both sides with respect to x, we get,

=> 4 (dy/dx) + 2x = 0

=> m₂ = dy/dx = −2x/4 =−x/2

For x = 2 and y = 1, we have m₁ = 2/2 = 1 and m₂ = −x/2 = −1.

Two curves intersect orthogonally if m₁m₂ = –1

=> 1 × (–1) = –1

Therefore these two curves intersect orthogonally at (2, 1).

Hence proved.

(ii) x² = y and x³ + 6y = 7 intersect orthogonally at (1, 1).

Solution:

First curve is x² = y

Differentiating both sides with respect to x, we get,

=> 2x = dy/dx

=> m₁ = dy/dx = 2x

Second curve is x³ + 6y = 7

Differentiating both sides with respect to x, we get,

=> 3x² + 6 (dy/dx) = 0

=> m₂ = dy/dx = −3x²/6 =−x²/2

For x = 1 and y = 1, we have m₁ = 2(1) = 2 and m₂ = −(1)²/2 = −1/2.

Two curves intersect orthogonally if m₁m₂ = –1

=> 2 × (–1/2) = –1

Therefore these two curves intersect orthogonally at (1, 1).

Hence proved.

(iii) y² = 8x and 2x² + y² = 10 intersect orthogonally at (1, 2√2).

Solution:

First curve is y² = 8x

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 8

=> m₁ = dy/dx = 8/2y = 4/y

Second curve is 2x² + y² = 10

Differentiating both sides with respect to x, we get,

=> 4x + 2y (dy/dx) = 0

=> m₂ = dy/dx = −4x/2y =−2x/y

For x = 1 and y = 2√2, we have m₁ = 4/2√2 = √2 and m₂ = −2/2√2 = −1/√2

Two curves intersect orthogonally if m₁m₂ = –1

=> √2 × (−1/√2) = –1

Therefore these two curves intersect orthogonally at (1, 2√2).

Hence proved.

Question 4. Show that the curves 4x = y² and 4xy = k cut at right angles, if k² = 512.

Solution:

First curve is 4x = y² . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 4

=> m₁ = dy/dx = 4/2y = 2/y

Second curve is 4xy = k . . . . (2)

Differentiating both sides with respect to x, we get,

=> y + x (dy/dx) = 0

=> m₂ = dy/dx = −y/x

Solving (1) and (2), we get,

=> y³ = k

=> y = k^1/3

So, x = k^2/3/4

As the curves intersect cut at right angles so, m₁m₂ = –1

=> (2/y) × (−y/x) = –1

=> 2/x = 1

=> 8/k^2/3 = 1

=> k^2/3 = 8

=> k² = 512

Hence proved.

Question 5. Show that the curves 2x = y² and 2xy = k cut at right angles, if k² = 8.

Solution:

First curve is 2x = y² . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 2

=> m₁ = dy/dx = 2/2y = 1/y

Second curve is 2xy = k . . . . (2)

Differentiating both sides with respect to x, we get,

=> y + x (dy/dx) = 0

=> m₂ = dy/dx = −y/x

Solving (1) and (2), we get,

=> y³ = k

=> y = k^1/3

So, x = k^2/3/2

As the curves intersect cut at right angles so, m₁m₂ = –1

=> (1/y) × (−y/x) = –1

=> 1/x = 1

=> 2/k^2/3 = 1

=> k^2/3 = 2

=> k² = 8

Hence proved.

Question 6. Prove that the curves xy = 4 and x² + y² = 8 touch each other.

Solution:

We have,

xy = 4 . . . . (1)

x² + y² = 8 . . . . (2)

Solving (1) and (2), we get,

=> (4/y)² + y² = 8

=> y⁴ − 8y² + 16 = 0

=> (y² − 4)² = 0

=> y = ±2

And we get x = ±2.

Differentiating eq. (1) with respect to x, we get,

=> y + x (dy/dx) = 0

=> m₁ = dy/dx = −y/x

Differentiating eq. (2) with respect to x, we get,

=> 2x + 2y (dy/dx) = 0

=> dy/dx = −x/y

At x = 2 and y = 2, we have,

m₁ = −2/2 = −1 and also m₂ = −2/2 = −1. Therefore m₁ = m₂.

Also at x = −2 and y = −2, we have m₁ = m₂

So, we can say that the curves touch each other at (2, 2) and (−2, −2).

Hence proved.

Question 7. Prove that the curves y² = 4x and x² + y² − 6x + 1 = 0 touch each other at the point (1, 2).

Solution:

We have,

y² = 4x . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 4

=> m₁ = dy/dx = 2/y

Also we have,

x² + y² − 6x + 1 = 0 . . . . (2)

Differentiating both with respect to x, we get,

=> 2x + 2y (dy/dx) − 6 = 0

=> m₂ = dy/dx = (6−2x)/2y = (3−x)/y

At x = 1 and y = 2, we have,

m₁ = 2/2 = 1

m₂ = (3−1)/2 = 1.

As m₁ = m₂, we can say that the curves touch each other at (1, 2).

Hence proved.

Question 8. Find the condition for the following curves to intersect orthogonally:

(i) x²/a² − y²/b² = 1 and xy = c²

Solution:

We have,

x²/a² − y²/b² = 1

Differentiating both sides with respect to x, we get,

=> 2x/a² − (2y/b²) (dy/dx) = 0

=> m₁ = dy/dx = b²x/a²y

Also, xy = c²

Differentiating both sides with respect to x, we get,

=> y + x (dy/dx) = 0

=> m₂ = dy/dx = −y/x

For curves to intersect orthogonally, m₁ m₂ = −1.

=> (b²x/a²y) (−y/x) = −1

=> a² = b²

Therefore, a² = b² is the condition for the curves to intersect orthogonally.

(ii) x²/a² + y²/b² = 1 and x²/A² − y²/B² = 1

Solution:

We have,

x²/a² + y²/b² = 1 . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x/a² + (2y/b²) (dy/dx) = 0

=> m₁ = dy/dx = −b²x/a²y

Also, x²/A² − y²/B² = 1 . . . . (2)

=> 2x/A² − (2y/B²) (dy/dx) = 0

=> m₂ = dy/dx = B²x/A²y

For curves to intersect orthogonally, m₁ m₂ = −1.

=> (−b²x/a²y) (B²x/A²y) = −1

=> x²/y² = a²A²/b²B² . . . . (3)

Subtracting (2) from (1) gives,

=>x^2[\frac{1}{a^2}-\frac{1}{A^2}]+y^2[\frac{1}{b^2}+\frac{1}{B^2}]=0

=>\frac{x^2}{y^2}=(\frac{b^2+B^2}{b^2B^2})×(\frac{a^2-A^2}{a^2A^2})

Putting this value in (3), we get,

=>\frac{b^2+B^2}{b^2B^2}×\frac{a^2-A^2}{a^2A^2}=\frac{a^2A^2}{b^2B^2}

=> B² + b² = a² − A²

=> a² − b² = A² + B²

Therefore, a² − b² = A² + B² is the condition for the curves to intersect orthogonally.

Question 9. Show that the curves\frac{x^2}{a^2+λ_1}+\frac{y^2}{b^2+λ_1}=1 and\frac{x^2}{a^2+λ_2}+\frac{y^2}{b^2+λ_2}=1 intersect at right angles.

Solution:

We have,

\frac{x^2}{a^2+λ_1}+\frac{y^2}{b^2+λ_1}=1 . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x/(a² + λ₁) + 2y/(b² + λ₁) (dy/dx) = 0

=> m₁ = dy/dx =\frac{−x(b^2 + λ_1)}{y(a^2 + λ_1)}

Also we have,

\frac{x^2}{a^2+λ_2}+\frac{y^2}{b^2+λ_2}=1 . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x/(a² + λ₂) + 2y/(b² + λ₂) (dy/dx) = 0

=> m₂ = dy/dx =\frac{−x(b^2 + λ_2)}{y(a^2 + λ_2)}

For curves to intersect orthogonally, m₁ m₂ = −1.

=>m_1m_2=\frac{−x(b^2 + λ_1)}{y(a^2 + λ_1)}×\frac{−x(b^2 + λ_2)}{y(a^2 + λ_2)}

=>m_1m_2=\frac{x^2}{y^2}×\frac{(b^2 + λ_1)(b^2 + λ_2)}{(a^2 + λ_1)(a^2 + λ_2)} . . . . (3)

Subtracting (2) from (1) gives,

=>x^2[\frac{1}{a^2+λ_1}-\frac{1}{a^2+λ_2}]+y^2[\frac{1}{b^2+λ_1}-\frac{1}{b^2+λ_2}]=0

=>\frac{x^2}{y^2}=\frac{λ_2-λ_1}{(b^2+λ_1)(b^2+λ_2)}×\frac{(a^2+λ_1)(a^2+λ_2)}{λ_1-λ_2}

Putting this value in (3), we get,

=>m_1m_2=\frac{λ_2-λ_1}{(b^2+λ_1)(b^2+λ_2)}×\frac{(a^2+λ_1)(a^2+λ_2)}{λ_1-λ_2}×\frac{(b^2 + λ_1)(b^2 + λ_2)}{(a^2 + λ_1)(a^2 + λ_2)}

=> m₁m₂ = −1

Hence proved.

Question 10. If the straight line x cos α + y sin α = p touches the curve x²/a² − y²/b² = 1, then prove that a² cos²α − b² sin²α = p².

Solution:

Suppose (x₁, y₁) is the point where the straight line x cos α + y sin α = p touches the curve

x²/a² − y²/b² = 1.

Now equation of tangent to x²/a² − y²/b² = 1 at (x₁, y₁) will be,

=>\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1

Therefore, the equation\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1 and the straight line x cos α + y sin α = p represent the same line. So, we get,

=>\frac{x_1}{a^2cosα}=\frac{-y_1}{b^2sinα}=\frac{1}{p}

=> x₁ = a² (cos α)/p and x₂ = b² (sin α)/p . . . . (1)

Now the point (x₁, y₁) lies on the curve x²/a² − y²/b² = 1.

=>\frac{x^2_1}{a^2}−\frac{y^2_1}{b^2}=1

Using (1), we get,

=>\frac{a^4cos^2α}{p^2a^2}-(\frac{-b^4sin^2α}{p^2b^2})=1

=> a² cos²α − b² sin²α = p²

Hence proved.

Summary

Exercise 16.3 in Chapter 16 of RD Sharma's Class 12 mathematics textbook focuses on advanced applications of tangents and normals to curves. This section challenges students to apply their knowledge of calculus, coordinate geometry, and algebraic manipulation to solve complex problems. The exercises typically cover a wide range of topics, including finding the length of tangents drawn from external points, determining points of contact for tangents with specific properties, calculating angles between normals and axes, and working with various curve types such as circles, parabolas, ellipses, and transcendental functions. Students are expected to demonstrate proficiency in calculating derivatives, manipulating equations, and interpreting geometric relationships. These problems not only reinforce fundamental concepts of tangents and normals but also encourage critical thinking and problem-solving skills, preparing students for more advanced mathematical analysis in higher education and real-world applications.