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

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter16
Exercise16.3
CategoryRD 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,

m= 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

m= −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 y= 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

=> m= 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 + y= 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 = k1/3

So, x = k2/3/4

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

=> (2/y) × (−y/x) = –1

=> 2/x = 1

=> 8/k2/3 = 1

=> k2/3 = 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 = k1/3

So, x = k2/3/2

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

=> (1/y) × (−y/x) = –1

=> 1/x = 1

=> 2/k2/3 = 1

=> k2/3 = 2

=> k2 = 8

Hence proved.

Question 6. Prove that the curves xy = 4 and x+ 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 = a(cos α)/p and x2 = b(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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