Here we provide RD Sharma Class 12 Ex 16.3 Solutions Chapter 16 Tangents and Normals for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 16.3 Solutions Chapter 16 Tangents and Normals book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 16 |

Exercise | 16.3 |

Category | RD Sharma Solutions |

**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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