RD Sharma Class 12 Ex 16.3 Solutions Chapter 16 Tangents and Normals

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	16
Exercise	16.3
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 16.3 Solutions Chapter 16 Tangents and Normals

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.

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RD Sharma Class 12 Ex 16.3 Solutions Chapter 16 Tangents and Normals

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

(i) y2 = x and x2 = y

(ii) y = x2 and x2 + y2 = 20

(iii) 2y2 = x3 and y2 = 32x

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

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

(vi) x2 + 4y2 = 8 and x2 – 2y2 = 2

(vii) x2 = 27y and y2 = 8x

(viii) x2 + y2 = 2x and y2 = x

(ix) y = 4 − x2 and y = x2

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

(i) y = x3 and 6y = 7 – x2

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

(iii) x2 + 4y2 = 8 and x2 – 2y2 = 4.

Question 3. Show that the curves:

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

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

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

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

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

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

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

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

(i) x2/a2 − y2/b2 = 1 and xy = c2

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

Question 9. Show that the curvesandintersect at right angles.

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.

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(i) y² = x and x² = y

(ii) y = x² and x² + y² = 20

(iii) 2y² = x³ and y² = 32x

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

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

(vi) x² + 4y² = 8 and x² – 2y² = 2

(vii) x² = 27y and y² = 8x

(viii) x² + y² = 2x and y²= x

(ix) y = 4 − x² and y = x²

(i) y = x³ and 6y = 7 – x²

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

(iii) x² + 4y² = 8 and x² – 2y² = 4.

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

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

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

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

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

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

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

(i) x²/a² − y²/b² = 1 and xy = c²

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

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.

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.