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

  • Last Updated : 28 Apr, 2021

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 = 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 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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