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

Last Updated : 28 Apr, 2021

### (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 âˆ’ x = 0

=> x (x3 âˆ’ 1) = 0

=> x = 0 or x = 1

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

tan Î¸ =

where m1 and m2 are the slopes of the curves.

When x = 0, then y = 0.

So, m1 = 1/2y = 1/0 = âˆž

m2 = 2x = 2(0) = 0

Therefore, tan Î¸ == âˆž

=> Î¸ = Ï€/2

When x = 1, then y = 1.

So, m1 = 1/2y = 1/2

m2 = 2x = 2(1) = 2

Therefore, tan Î¸ =

=> Î¸ = tanâˆ’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 = âˆ’x/y

Solving (1) and (2), we get,

=> y2 +y âˆ’ 20 = 0

=> y2 + 5y âˆ’ 4y âˆ’ 20 = 0

=> (y + 5) (y âˆ’ 4) = 0

=> y = âˆ’5 or y = 4

Ignoring y = âˆ’ 5 as x becomes âˆš(âˆ’5) in that case, which is not possible.

When y = 4, we get x2 = 4

=> x = Â±2

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

tan Î¸ =

where m1 and m2 are the slopes of the curves.

When x = Â±2 and y = 4, we get,

m1 = 2x = 2(2) = 4 or Â±4

m2 = âˆ’x/y = âˆ’2/4 = âˆ’1/2

So, tan Î¸ =

=> Î¸ = tanâˆ’1 (9/2)

When x = âˆ’2 and y = 4, we get,

m1 = 2x = 4 or âˆ’4

m2 = âˆ’x/y = 1/2 or âˆ’1/2

So, tan Î¸ =

=> Î¸ = tanâˆ’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 âˆ’ 64x = 0

=> x(x2 âˆ’ 64) = 0

=> x = 0 or x2 âˆ’ 64 = 0

=> x = 0 or x = Â±8

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

tan Î¸ =

where m1 and m2 are the slopes of the curves.

When x = 0 then y = 0.

m1 = 3x2/4y = âˆž

m2 = 16/y = âˆž

So, tan Î¸ = âˆž

=> Î¸ = Ï€/2

When x = Â±8, then y = Â±16.

m1 = 3x2/4y = 3 or âˆ’3

m2 = 16/y = 1 or âˆ’1

So, tan Î¸ =

=> Î¸ = tanâˆ’1 (1/2)

### (iv) x2 + y2 â€“ 4x â€“ 1 = 0 and x2 + y2 â€“ 2y â€“ 9 = 0

Solution:

First curve is x2 + y2 â€“ 4x â€“ 1 = 0. . . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) â€“ 4 = 0

=> m1 = dy/dx = (2â€“x)/y

Second curve is x2 + y2 â€“ 2y â€“ 9 = 0. . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x + 2y (dy/dx) â€“ 2 (dy/dx) = 0

=> m2 = dy/dx = â€“x/(yâ€“1)

First curve can be written as,

=> (x â€“ 2)2 + y2 â€“ 5 = 0 . . . . (3)

Subtracting (2) from (1), we get

=> x2 + y2 â€“ 4x â€“ 1 â€“ x2 â€“ y2 + 2y + 9 = 0

=> â€“ 4x â€“ 1 + 2y + 9 = 0

=> 2y = 4x â€“ 8

=> y = 2x â€“ 4

Putting y = 2x â€“ 4 in (1), we get,

=> (x â€“ 2)2 + (2x â€“ 4)2 â€“ 5 = 0

â‡’ (x â€“ 2)2(1 + 4) â€“ 5 = 0

â‡’ 5(x â€“ 2)2 â€“ 5 = 0

â‡’ (x â€“ 2)2 = 1

â‡’ x = 3 or x = 1

So, when x = 3 then y = 6 â€“ 4 = 2

m1 = (2â€“x)/y = (2â€“3)/2 = â€“1/2

m2 = â€“x/(yâ€“1) = â€“3/(2â€“1) = â€“3

So, tan Î¸ == 1

=> Î¸ = Ï€/4

So, when x = 1 then y = 2 â€“ 4 = â€“ 2

m1 = (2â€“x)/y = (2â€“1)/(â€“2) = â€“1/2

m2 = â€“x/(yâ€“1) = â€“1/(â€“2â€“1) = 1/3

So, tan Î¸ == 1

=> Î¸ = Ï€/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 = â€“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 = â€“2x/2y = â€“x/y

Solving (1) and (2), we get,

=> x2/a2 + (ab â€“ x2)/b2 = 1

=> x2b2 â€“ a2x2 = a2b2 â€“ a3b

=> x2 =

=> x =

From (2), we get, y2 =

=> y =

So, m1 = â€“b2x/a2y =

=

m2 = â€“x/y =

=

Therefore, tan Î¸ =

=> tan Î¸ =

=> tan Î¸ =

=> Î¸ = tanâ€“1 ((aâ€“b)/âˆšab)

### (vi) x2 + 4y2 = 8 and x2 â€“ 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 = â€“2x/8y = â€“x/4y

Second curve is x2 â€“ 2y2 = 2 . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x â€“ 4y (dy/dx) = 0

=> m2 = dy/dx = x/2y

Solving (1) and (2), we get,

6y2 = 6 => y2 = Â±1

x2 = 2 + 2 => x = Â±2

So, tan Î¸ =

=> Î¸ = tanâ€“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 âˆ’ 1728) = 0

=> y = 0 or y = 12

And x = 0 or x = 18.

So, when x = 0 and y = 0

m1 = 0 and m2 = âˆž

tan Î¸ == âˆž

=> Î¸ = Ï€/2

So, when x = 18 and y = 12

m1 = 2x/27 = 12/9 = 4/3 and m2 = 4/y = 1/3

tan Î¸ =

=> Î¸ = tanâˆ’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 = (1â€“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 â€“ x = 0

=> x = 0 or x = 1

And y = 0 or y = Â±1.

When x = 0, y = 0, m1 = âˆž and m2 = âˆž

tan Î¸ =

=> Î¸ = Ï€/2

When x = 1 and y = Â±1, m1 = 0 and m2 = 1/2

tan Î¸ =

=> Î¸ = tanâˆ’1 (1/2)

### (ix) y = 4 âˆ’ x2 and y = x2

Solution:

First curve is y = 4 âˆ’ x2 . . . . (1)

Differentiating both sides with respect to x, we get,

=> dy/dx = âˆ’2x

=> m1 = dy/dx = âˆ’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 = Â±âˆš2

And y = 2

So, m1 = âˆ’2x = âˆ’2âˆš2 and m2 = 2x = 2âˆš2

tan Î¸ =

=> Î¸ = tanâˆ’1 (4âˆš2/7)

### (i) y = x3 and 6y = 7 â€“ 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 â€“ x2 . . . . (2)

Differentiating both sides with respect to x, we get,

=> 6y (dy/dx) = â€“ 2x

=> m2 = dy/dx = â€“2x/6y = â€“ x/3y

Solving (1) and (2), we get,

=> 6y = 7 â€“ x2

=> 6x3 + x2 â€“ 7 = 0

As x = 1 satisfies this equation, we get x = 1 and y = 13 = 1

So, m1 = 3 and m2 = â€“ 1/3

Two curves intersect orthogonally if m1m2 = â€“1

=> 3 Ã— (â€“1/3) = â€“1

Hence proved.

### (ii) x3 â€“ 3xy2 = â€“ 2 and 3x2 y â€“ y3 = 2

Solution:

First curve is x3 â€“ 3xy2 = â€“ 2

Differentiating both sides with respect to x, we get,

=> 3x2 â€“ 3y2 â€“ 6xy (dy/dx) = 0

=> m1 = dy/dx = 3(x2â€“y2)/6xy

Second curve is 3x2y â€“ y3 = 2

Differentiating both sides with respect to x, we get,

=> 6xy + 3x2 (dy/dx) â€“ 3y2 (dy/dx) = 0

=> m2 = dy/dx = â€“6xy/3(x2â€“y2)

Two curves intersect orthogonally if m1m2 = â€“1

=>= â€“1

Hence proved.

### (iii) x2 + 4y2 = 8 and x2 â€“ 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 = â€“x/4y

Second curve is x2 â€“ 2y2 = 4 . . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x â€“ 4y (dy/dx) = 0

=> m2 = dy/dx = 2x/4y = x/2y

Solving (1) and (2), we get,

=> x = 4/âˆš3 and y = âˆš2/âˆš3

So, m1 = â€“x/4y = â€“1/âˆš2

m2 = x/2y = âˆš2

Two curves intersect orthogonally if m1m2 = â€“1

=> (â€“1/âˆš2) Ã— âˆš2 = â€“1

Hence proved.

### (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 = âˆ’2x/4 =âˆ’x/2

For x = 2 and y = 1, we have m1 = 2/2 = 1 and m2 = âˆ’x/2 = âˆ’1.

Two curves intersect orthogonally if m1m2 = â€“1

=> 1 Ã— (â€“1) = â€“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 = âˆ’3x2/6 =âˆ’x2/2

For x = 1 and y = 1, we have m1 = 2(1) = 2 and m2 = âˆ’(1)2/2 = âˆ’1/2.

Two curves intersect orthogonally if m1m2 = â€“1

=> 2 Ã— (â€“1/2) = â€“1

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

Hence proved.

### (iii) y2 = 8x and 2x2 + y2 = 10intersect orthogonallyat (1, 2âˆš2).

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 = âˆ’4x/2y =âˆ’2x/y

For x = 1 and y = 2âˆš2, we have m1 = 4/2âˆš2 = âˆš2 and m2 = âˆ’2/2âˆš2 = âˆ’1/âˆš2

Two curves intersect orthogonally if m1m2 = â€“1

=> âˆš2 Ã— (âˆ’1/âˆš2) = â€“1

Therefore these two curves intersect orthogonally at (1, 2âˆš2).

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 = âˆ’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 = â€“1

=> (2/y) Ã— (âˆ’y/x) = â€“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 = âˆ’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 = â€“1

=> (1/y) Ã— (âˆ’y/x) = â€“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 âˆ’ 8y2 + 16 = 0

=> (y2 âˆ’ 4)2 = 0

=> y = Â±2

And we get x = Â±2.

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

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

=> m1 = dy/dx = âˆ’y/x

Differentiating eq. (2) with respect to x, we get,

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

=> dy/dx = âˆ’x/y

At x = 2 and y = 2, we have,

m1 = âˆ’2/2 = âˆ’1 and also m2 = âˆ’2/2 = âˆ’1. Therefore m1 = m2.

Also at x = âˆ’2 and y = âˆ’2, we have m1 = m2

So, we can say that the curves touch each other at (2, 2) and (âˆ’2, âˆ’2).

Hence proved.

### Question 7. Prove that the curves y2 = 4x and x2 + y2 âˆ’ 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 âˆ’ 6x + 1 = 0 . . . . (2)

Differentiating both with respect to x, we get,

=> 2x + 2y (dy/dx) âˆ’ 6 = 0

=> m2 = dy/dx = (6âˆ’2x)/2y = (3âˆ’x)/y

At x = 1 and y = 2, we have,

m1 = 2/2 = 1

m2 = (3âˆ’1)/2 = 1.

As m1 = m2, we can say that the curves touch each other at (1, 2).

Hence proved.

### (i) x2/a2 âˆ’ y2/b2 = 1 and xy = c2

Solution:

We have,

x2/a2 âˆ’ y2/b2 = 1

Differentiating both sides with respect to x, we get,

=> 2x/a2 âˆ’ (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 = âˆ’y/x

For curves to intersect orthogonally, m1 m2 = âˆ’1.

=> (b2x/a2y) (âˆ’y/x) = âˆ’1

=> a2 = b2

Therefore, a2 = b2 is the condition for the curves to intersect orthogonally.

### (ii) x2/a2 + y2/b2 = 1 and x2/A2 âˆ’ 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 = âˆ’b2x/a2y

Also, x2/A2 âˆ’ y2/B2 = 1 . . . . (2)

=> 2x/A2 âˆ’ (2y/B2) (dy/dx) = 0

=> m2 = dy/dx = B2x/A2y

For curves to intersect orthogonally, m1 m2 = âˆ’1.

=> (âˆ’b2x/a2y) (B2x/A2y) = âˆ’1

=> x2/y2 = a2A2/b2B2 . . . . (3)

Subtracting (2) from (1) gives,

=>

=>

Putting this value in (3), we get,

=>

=> B2 + b2 = a2 âˆ’ A2

=> a2 âˆ’ b2 = A2 + B2

Therefore, a2 âˆ’ b2 = A2 + B2 is the condition for the curves to intersect orthogonally.

### Question 9. Show that the curvesandintersect at right angles.

Solution:

We have,

. . . . (1)

Differentiating both sides with respect to x, we get,

=> 2x/(a2 + Î»1) + 2y/(b2 + Î»1) (dy/dx) = 0

=> m1 = dy/dx =

Also we have,

. . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x/(a2 + Î»2) + 2y/(b2 + Î»2) (dy/dx) = 0

=> m2 = dy/dx =

For curves to intersect orthogonally, m1 m2 = âˆ’1.

=>

=>. . . . (3)

Subtracting (2) from (1) gives,

=>

=>

Putting this value in (3), we get,

=>

=> m1m2 = âˆ’1

Hence proved.

### Question 10. If the straight line x cos Î± + y sin Î± = p touches the curve x2/a2 âˆ’ y2/b2 = 1, then prove that a2 cos2Î± âˆ’ b2 sin2Î± = p2.

Solution:

Suppose (x1, y1) is the point where the straight line x cos Î± + y sin Î± = p touches the curve

x2/a2 âˆ’ y2/b2 = 1.

Now equation of tangent to x2/a2 âˆ’ y2/b2 = 1 at (x1, y1) will be,

=>

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

=>

=> x1 = a2 (cos Î±)/p and x2 = b2 (sin Î±)/p . . . . (1)

Now the point (x1, y1) lies on the curve x2/a2 âˆ’ y2/b2 = 1.

=>

Using (1), we get,

=>

=> a2 cos2Î± âˆ’ b2 sin2Î± = p2

Hence proved.

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