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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) y**^{2} = x and x^{2} = y

^{2}= x and x

^{2}= y

**Solution:**

First curve is y

^{2}= x. . . . . (1)Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 1

=> m

_{1}= dy/dx = 1/2ySecond curve is x

^{2}= y . . . . (2)Differentiating both sides with respect to x, we get,

=> 2x = dy/dx

=> m

_{2}= dy/dx = 2xSolving (1) and (2), we get,

=> x

^{4}− x = 0=> x (x

^{3}− 1) = 0=> x = 0 or x = 1

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

tan θ =

where m

_{1}and m_{2}are the slopes of the curves.When x = 0, then y = 0.

So, m

_{1}= 1/2y = 1/0 = ∞m

_{2}= 2x = 2(0) = 0Therefore, tan θ == ∞

=> θ = π/2

When x = 1, then y = 1.

So, m

_{1}= 1/2y = 1/2m

_{2}= 2x = 2(1) = 2Therefore, tan θ =

=> θ = tan

^{−1 }(3/4)

**(ii) y = x**^{2} and x^{2} + y^{2} = 20

^{2}and x

^{2}+ y

^{2}= 20

**Solution:**

First curve is y = x

^{2}. . . . . (1)Differentiating both sides with respect to x, we get,

=> (dy/dx) = 2x

=> m

_{1}= dy/dx = 2xSecond curve is x

^{2}+ y^{2}= 20 . . . . (2)Differentiating both sides with respect to x, we get,

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

=> m

_{2}= dy/dx = −x/ySolving (1) and (2), we get,

=> y

^{2}+y − 20 = 0=> y

^{2}+ 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 x

^{2}= 4=> x = ±2

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

tan θ =

where m

_{1}and m_{2}are the slopes of the curves.When x = ±2 and y = 4, we get,

m

_{1 }= 2x = 2(2) = 4 or ±4m

_{2}= −x/y = −2/4 = −1/2So, tan θ =

=> θ = tan

^{−1}(9/2)When x = −2 and y = 4, we get,

m

_{1}= 2x = 4 or −4m

_{2 }= −x/y = 1/2 or −1/2So, tan θ =

=> θ = tan

^{−1}(9/2)

**(iii) 2y**^{2} = x^{3} and y^{2} = 32x

^{2}= x

^{3}and y

^{2}= 32x

**Solution:**

First curve is 2y

^{2}= x^{3}. . . . . (1)Differentiating both sides with respect to x, we get,

=> 4y (dy/dx) = 3x

^{2}=> m

_{1}= dy/dx = 3x^{2}/4ySecond curve is y

^{2}= 32x. . . . . (2)Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 32

=> m

_{2}= dy/dx = 32/2y = 16/ySolving (1) and (2), we get,

=> 2(32x) = x

^{3}=> x

^{3}− 64x = 0=> x(x

^{2}− 64) = 0=> x = 0 or x

^{2}− 64 = 0=> x = 0 or x = ±8

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

tan θ =

where m

_{1}and m_{2}are the slopes of the curves.When x = 0 then y = 0.

m

_{1}= 3x^{2}/4y = ∞m

_{2}= 16/y = ∞So, tan θ = ∞

=> θ = π/2

When x = ±8, then y = ±16.

m

_{1}= 3x^{2}/4y = 3 or −3m

_{2}= 16/y = 1 or −1So, tan θ =

=> θ = tan

^{−1}(1/2)

**(iv) x**^{2} + y^{2} – 4x – 1 = 0 and x^{2} + y^{2} – 2y – 9 = 0

^{2}+ y

^{2}– 4x – 1 = 0 and x

^{2}+ y

^{2}– 2y – 9 = 0

**Solution:**

First curve is x

^{2}+ y^{2}– 4x – 1 = 0. . . . . (1)Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) – 4 = 0

=> m

_{1}= dy/dx = (2–x)/ySecond curve is x

^{2}+ y^{2}– 2y – 9 = 0. . . . . (2)Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) – 2 (dy/dx) = 0

=> m

_{2}= dy/dx = –x/(y–1)First curve can be written as,

=> (x – 2)

^{2}+ y^{2}– 5 = 0 . . . . (3)Subtracting (2) from (1), we get

=> x

^{2}+ y^{2}– 4x – 1 – x^{2}– y^{2}+ 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

m

_{1}= (2–x)/y = (2–3)/2 = –1/2m

_{2}= –x/(y–1) = –3/(2–1) = –3So, tan θ == 1

=> θ = π/4

So, when x = 1 then y = 2 – 4 = – 2

m

_{1}= (2–x)/y = (2–1)/(–2) = –1/2m

_{2}= –x/(y–1) = –1/(–2–1) = 1/3So, tan θ == 1

=> θ = π/4

**(v) x**^{2}/a^{2} + y^{2}/b^{2} = 1 and x^{2} + y^{2} = ab

^{2}/a

^{2}+ y

^{2}/b

^{2}= 1 and x

^{2}+ y

^{2}= ab

**Solution:**

First curve is x

^{2}/a^{2}+ y^{2}/b^{2}= 1 . . . . (1)Differentiating both sides with respect to x, we get,

=> 2x/a

^{2}+ (2y/b^{2}) (dy/dx) = 0=> m

_{1}= dy/dx = –b^{2}x/a^{2}ySecond curve is x

^{2}+ y^{2}= ab . . . . (2)Differentiating both sides with respect to x, we get,

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

=> m

_{2}= dy/dx = –2x/2y = –x/ySolving (1) and (2), we get,

=> x

^{2}/a^{2}+ (ab – x^{2})/b^{2}= 1=> x

^{2}b^{2}– a^{2}x^{2}= a^{2}b^{2}– a^{3}b=> x

^{2}==> x =

From (2), we get, y

^{2}==> y =

So, m

_{1}= –b^{2}x/a^{2}y ==

m

_{2}= –x/y ==

Therefore, tan θ =

=> tan θ =

=> tan θ =

=> θ = tan

^{–1}((a–b)/√ab)

**(vi) x**^{2} + 4y^{2} = 8 and x^{2} – 2y^{2} = 2

^{2}+ 4y

^{2}= 8 and x

^{2}– 2y

^{2}= 2

**Solution:**

First curve is x

^{2}+ 4y^{2}= 8 . . . . (1)Differentiating both sides with respect to x, we get,

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

=> m

_{1}= dy/dx = –2x/8y = –x/4ySecond curve is x

^{2}– 2y^{2}= 2 . . . . (2)Differentiating both sides with respect to x, we get,

=> 2x – 4y (dy/dx) = 0

=> m

_{2}= dy/dx = x/2ySolving (1) and (2), we get,

6y

^{2}= 6 => y2 = ±1x

^{2}= 2 + 2 => x = ±2So, tan θ =

=> θ = tan

^{–1}(1/3)

**(vii) x**^{2} = 27y and y^{2} = 8x

^{2}= 27y and y

^{2}= 8x

**Solution:**

First curve is x

^{2}= 27y . . . . (1)Differentiating both sides with respect to x, we get,

=> 2x = 27 (dy/dx)

=> m

_{1}= dy/dx = 2x/27Second curve is y

^{2}= 8x . . . . (2)Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 8

=> m

_{2}= dy/dx = 8/2y = 4/ySolving (1) and (2), we get,

=> y

^{4}/64 = 27y=> y (y

^{3}− 1728) = 0=> y = 0 or y = 12

And x = 0 or x = 18.

So, when x = 0 and y = 0

m

_{1}= 0 and m_{2}= ∞tan θ == ∞

=> θ = π/2

So, when x = 18 and y = 12

m

_{1}= 2x/27 = 12/9 = 4/3 and m_{2}= 4/y = 1/3tan θ =

=> θ = tan

^{−1}(9/13)

**(viii) x**^{2} + y^{2} = 2x and y^{2 }= x

^{2}+ y

^{2}= 2x and y

^{2 }= x

**Solution:**

First curve is x

^{2}+ y^{2}= 2x . . . . (1)Differentiating both sides with respect to x, we get,

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

=> m

_{1}= dy/dx = (1–x)/ySecond curve is y

^{2}= x . . . . (2)Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 1

=> m

_{2}= dy/dx = 1/2ySolving (1) and (2), we get,

=> x

^{2}– x = 0=> x = 0 or x = 1

And y = 0 or y = ±1.

When x = 0, y = 0, m

_{1}= ∞ and m_{2}= ∞tan θ =

=> θ = π/2

When x = 1 and y = ±1, m

_{1}= 0 and m_{2}= 1/2tan θ =

=> θ = tan

^{−1}(1/2)

**(ix) y = 4 − x**^{2} and y = x^{2}

^{2}and y = x

^{2}

**Solution:**

First curve is y = 4 − x

^{2}. . . . (1)Differentiating both sides with respect to x, we get,

=> dy/dx = −2x

=> m

_{1}= dy/dx = −2xSecond curve is y = x

^{2}. . . . (2)Differentiating both sides with respect to x, we get,

=> dy/dx = 2x

=> m

_{2 }= dy/dx = 2xSolving (1) and (2), we get,

=> 2x

^{2}= 4=> x = ±√2

And y = 2

So, m

_{1}= −2x = −2√2 and m_{2}= 2x = 2√2tan θ =

=> θ = tan

^{−1}(4√2/7)

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

**(i) y = x**^{3} and 6y = 7 – x^{2}

^{3}and 6y = 7 – x

^{2}

**Solution:**

First curve is y = x

^{3}. . . . (1)Differentiating both sides with respect to x, we get,

=> dy/dx = 3x

^{2}=> m

_{1}= dy/dx = 3x^{2}Second curve is 6y = 7 – x

^{2}. . . . (2)Differentiating both sides with respect to x, we get,

=> 6y (dy/dx) = – 2x

=> m

_{2}= dy/dx = –2x/6y = – x/3ySolving (1) and (2), we get,

=> 6y = 7 – x

^{2}=> 6x

^{3}+ x^{2}– 7 = 0As x = 1 satisfies this equation, we get x = 1 and y = 1

^{3}= 1So, m

_{1}= 3 and m_{2}= – 1/3Two curves intersect orthogonally if m

_{1}m_{2}= –1=> 3 × (–1/3) = –1

Hence proved.

**(ii) x**^{3} – 3xy^{2} = – 2 and 3x^{2} y – y^{3} = 2

^{3}– 3xy

^{2}= – 2 and 3x

^{2}y – y

^{3}= 2

**Solution:**

First curve is x

^{3}– 3xy^{2}= – 2Differentiating both sides with respect to x, we get,

=> 3x

^{2}– 3y^{2}– 6xy (dy/dx) = 0=> m

_{1}= dy/dx = 3(x^{2}–y^{2})/6xySecond curve is 3x

^{2}y – y^{3}= 2Differentiating both sides with respect to x, we get,

=> 6xy + 3x

^{2}(dy/dx) – 3y^{2}(dy/dx) = 0=> m

_{2}= dy/dx = –6xy/3(x^{2}–y^{2})Two curves intersect orthogonally if m

_{1}m_{2}= –1=>= –1

Hence proved.

**(iii) x**^{2} + 4y^{2} = 8 and x^{2} – 2y^{2} = 4.

^{2}+ 4y

^{2}= 8 and x

^{2}– 2y

^{2}= 4.

**Solution:**

First curve is x

^{2}+ 4y^{2}= 8 . . . . (1)Differentiating both sides with respect to x, we get,

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

=> m

_{1}= dy/dx = 2x/8y = –x/4ySecond curve is x

^{2}– 2y^{2}= 4 . . . . (2)Differentiating both sides with respect to x, we get,

=> 2x – 4y (dy/dx) = 0

=> m

_{2}= dy/dx = 2x/4y = x/2ySolving (1) and (2), we get,

=> x = 4/√3 and y = √2/√3

So, m

_{1}= –x/4y = –1/√2m

_{2}= x/2y = √2Two curves intersect orthogonally if m

_{1}m_{2}= –1=> (–1/√2) × √2 = –1

Hence proved.

**Question 3. Show that the curves:**

**(i) x**^{2} = 4y and 4y + x^{2} = 8 intersect orthogonally at (2, 1).

^{2}= 4y and 4y + x

^{2}= 8 intersect orthogonally at (2, 1).

**Solution:**

First curve is x

^{2}= 4yDifferentiating both sides with respect to x, we get,

=> 2x = 4 (dy/dx)

=> m

_{1}= dy/dx = 2x/4 = x/2Second curve is 4y + x

^{2}= 8Differentiating both sides with respect to x, we get,

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

=> m

_{2}= dy/dx = −2x/4 =−x/2For x = 2 and y = 1, we have m

_{1}= 2/2 = 1 and m_{2}= −x/2 = −1.Two curves intersect orthogonally if m

_{1}m_{2}= –1=> 1 × (–1) = –1

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

Hence proved.

**(ii) x**^{2} = y and x^{3} + 6y = 7 intersect orthogonally at (1, 1).

^{2}= y and x

^{3}+ 6y = 7 intersect orthogonally at (1, 1).

**Solution:**

First curve is x

^{2}= yDifferentiating both sides with respect to x, we get,

=> 2x = dy/dx

=> m

_{1}= dy/dx = 2xSecond curve is x

^{3}+ 6y = 7Differentiating both sides with respect to x, we get,

=> 3x

^{2}+ 6 (dy/dx) = 0=> m

_{2}= dy/dx = −3x^{2}/6 =−x^{2}/2For x = 1 and y = 1, we have m

_{1}= 2(1) = 2 and m_{2}= −(1)^{2}/2 = −1/2.Two curves intersect orthogonally if m

_{1}m_{2}= –1=> 2 × (–1/2) = –1

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

Hence proved.

**(iii) y**^{2} = 8x and 2x^{2} + y^{2} = 10 **intersect orthogonally** **at (1, 2√2).**

^{2}= 8x and 2x

^{2}+ y

^{2}= 10

**Solution:**

First curve is y

^{2}= 8xDifferentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 8

=> m

_{1}= dy/dx = 8/2y = 4/ySecond curve is 2x

^{2}+ y^{2 }= 10Differentiating both sides with respect to x, we get,

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

=> m

_{2}= dy/dx = −4x/2y =−2x/yFor x = 1 and y = 2√2, we have m

_{1}= 4/2√2 = √2 and m_{2}= −2/2√2 = −1/√2Two curves intersect orthogonally if m

_{1}m_{2}= –1=> √2 × (−1/√2) = –1

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

Hence proved.

**Question 4. Show that the curves 4x = y**^{2} and 4xy = k cut at right angles, if k^{2} = 512.

^{2}and 4xy = k cut at right angles, if k

^{2}= 512.

**Solution:**

First curve is 4x = y

^{2}. . . . (1)Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 4

=> m

_{1}= dy/dx = 4/2y = 2/ySecond curve is 4xy = k . . . . (2)

Differentiating both sides with respect to x, we get,

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

=> m

_{2}= dy/dx = −y/xSolving (1) and (2), we get,

=> y

^{3}= k=> y = k

^{1/3}So, x = k

^{2/3}/4As the curves intersect cut at right angles so, m

_{1}m_{2}= –1=> (2/y) × (−y/x) = –1

=> 2/x = 1

=> 8/k

^{2/3}= 1=> k

^{2/3}= 8=> k

^{2}= 512

Hence proved.

**Question 5. Show that the curves 2x = y**^{2} and 2xy = k cut at right angles, if k^{2} = 8.

^{2}and 2xy = k cut at right angles, if k

^{2}= 8.

**Solution:**

First curve is 2x = y

^{2}. . . . (1)Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 2

=> m

_{1}= dy/dx = 2/2y = 1/ySecond curve is 2xy = k . . . . (2)

Differentiating both sides with respect to x, we get,

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

=> m

_{2}= dy/dx = −y/xSolving (1) and (2), we get,

=> y

^{3}= k=> y = k

^{1/3}So, x = k

^{2/3}/2As the curves intersect cut at right angles so, m

_{1}m_{2}= –1=> (1/y) × (−y/x) = –1

=> 1/x = 1

=> 2/k

^{2/3}= 1=> k

^{2/3}= 2=> k

^{2}= 8

Hence proved.

**Question 6. Prove that the curves xy = 4 and x**^{2 }+ y^{2} = 8 touch each other.

^{2 }+ y

^{2}= 8 touch each other.

**Solution:**

We have,

xy = 4 . . . . (1)

x

^{2}+ y^{2}= 8 . . . . (2)Solving (1) and (2), we get,

=> (4/y)

^{2}+ y^{2}= 8=> y

^{4}− 8y^{2}+ 16 = 0=> (y

^{2}− 4)^{2}= 0=> y = ±2

And we get x = ±2.

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

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

=> m

_{1}= dy/dx = −y/xDifferentiating 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,

m

_{1}= −2/2 = −1 and also m_{2}= −2/2 = −1. Therefore m_{1}= m_{2}.Also at x = −2 and y = −2, we have m

_{1}= m_{2}So, we can say that the curves touch each other at (2, 2) and (−2, −2).

Hence proved.

**Question 7. Prove that the curves y**^{2} = 4x and x^{2} + y^{2} − 6x + 1 = 0 touch each other at the point (1, 2).

^{2}= 4x and x

^{2}+ y

^{2}− 6x + 1 = 0 touch each other at the point (1, 2).

**Solution:**

We have,

y

^{2}= 4x . . . . (1)Differentiating both sides with respect to x, we get,

=> 2y (dy/dx) = 4

=> m

_{1}= dy/dx = 2/yAlso we have,

x

^{2}+ y^{2}− 6x + 1 = 0 . . . . (2)Differentiating both with respect to x, we get,

=> 2x + 2y (dy/dx) − 6 = 0

=> m

_{2}= dy/dx = (6−2x)/2y = (3−x)/yAt x = 1 and y = 2, we have,

m

_{1}= 2/2 = 1m

_{2}= (3−1)/2 = 1.As m

_{1}= m_{2}, 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) x**^{2}/a^{2} − y^{2}/b^{2} = 1 and xy = c^{2}

^{2}/a

^{2}− y

^{2}/b

^{2}= 1 and xy = c

^{2}

**Solution:**

We have,

x

^{2}/a^{2}− y^{2}/b^{2}= 1Differentiating both sides with respect to x, we get,

=> 2x/a

^{2}− (2y/b^{2}) (dy/dx) = 0=> m

_{1}= dy/dx = b^{2}x/a^{2}yAlso, xy = c

^{2}Differentiating both sides with respect to x, we get,

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

=> m

_{2}= dy/dx = −y/xFor curves to intersect orthogonally, m

_{1}m_{2}= −1.=> (b

^{2}x/a^{2}y) (−y/x) = −1=> a

^{2}= b^{2}

Therefore, a^{2}= b^{2}is the condition for the curves to intersect orthogonally.

**(ii) x**^{2}/a^{2} + y^{2}/b^{2} = 1 and x^{2}/A^{2} − y^{2}/B^{2} = 1

^{2}/a

^{2}+ y

^{2}/b

^{2}= 1 and x

^{2}/A

^{2}− y

^{2}/B

^{2}= 1

**Solution:**

We have,

x

^{2}/a^{2}+ y^{2}/b^{2}= 1 . . . . (1)Differentiating both sides with respect to x, we get,

=> 2x/a

^{2}+ (2y/b^{2}) (dy/dx) = 0=> m

_{1}= dy/dx = −b^{2}x/a^{2}yAlso, x

^{2}/A^{2}− y^{2}/B^{2}= 1 . . . . (2)=> 2x/A

^{2}− (2y/B^{2}) (dy/dx) = 0=> m

_{2}= dy/dx = B^{2}x/A^{2}yFor curves to intersect orthogonally, m

_{1}m_{2}= −1.=> (−b

^{2}x/a^{2}y) (B^{2}x/A^{2}y) = −1=> x

^{2}/y^{2}= a^{2}A^{2}/b^{2}B^{2}. . . . (3)Subtracting (2) from (1) gives,

=>

=>

Putting this value in (3), we get,

=>

=> B

^{2}+ b^{2}= a^{2}− A^{2}=> a

^{2}− b^{2}= A^{2}+ B^{2}

Therefore, a^{2}− b^{2}= A^{2}+ B^{2}is the condition for the curves to intersect orthogonally.

**Question 9. Show that the curves****and****intersect at right angles.**

**Solution:**

We have,

. . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x/(a

^{2}+ λ_{1}) + 2y/(b^{2}+ λ_{1}) (dy/dx) = 0=> m

_{1}= dy/dx =Also we have,

. . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x/(a

^{2}+ λ_{2}) + 2y/(b^{2}+ λ_{2}) (dy/dx) = 0=> m

_{2}= dy/dx =For curves to intersect orthogonally, m

_{1}m_{2}= −1.=>

=>. . . . (3)

Subtracting (2) from (1) gives,

=>

=>

Putting this value in (3), we get,

=>

=> m

_{1}m_{2}= −1

Hence proved.

**Question 10. If the straight line x cos α + y sin α = p touches the curve x**^{2}/a^{2} − y^{2}/b^{2} = 1, then prove that a^{2} cos^{2}α − b^{2} sin^{2}α = p^{2}.

^{2}/a

^{2}− y

^{2}/b

^{2}= 1, then prove that a

^{2}cos

^{2}α − b

^{2}sin

^{2}α = p

^{2}.

**Solution:**

Suppose (x

_{1}, y_{1}) is the point where the straight line x cos α + y sin α = p touches the curvex

^{2}/a^{2}− y^{2}/b^{2}= 1.Now equation of tangent to x

^{2}/a^{2}− y^{2}/b^{2}= 1 at (x_{1}, y_{1}) will be,=>

Therefore, the equationand the straight line x cos α + y sin α = p represent the same line. So, we get,

=>

=> x

_{1}= a^{2 }(cos α)/p and x_{2}= b^{2 }(sin α)/p . . . . (1)Now the point (x

_{1}, y_{1}) lies on the curve x^{2}/a^{2}− y^{2}/b^{2}= 1.=>

Using (1), we get,

=>

=> a

^{2}cos^{2}α − b^{2}sin^{2}α = p^{2}

Hence proved.

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