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

**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}âˆ’ x = 0=> x (x

^{3}âˆ’ 1) = 0=> x = 0 or x = 1

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

tan Î¸ =

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 = âˆžm

_{2}= 2x = 2(0) = 0Therefore, tan Î¸ == âˆž

=> Î¸ = Ï€/2

When x = 1, then y = 1.

So, m

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

_{2}= 2x = 2(1) = 2Therefore, tan Î¸ =

=> Î¸ = tan

^{âˆ’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 = âˆ’x/ySolving (1) and (2), we get,

=> y

^{2}+y âˆ’ 20 = 0=> y

^{2}+ 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 x

^{2}= 4=> x = Â±2

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

tan Î¸ =

where m

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

m

_{1 }= 2x = 2(2) = 4 or Â±4m

_{2}= âˆ’x/y = âˆ’2/4 = âˆ’1/2So, tan Î¸ =

=> Î¸ = tan

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

m

_{1}= 2x = 4 or âˆ’4m

_{2 }= âˆ’x/y = 1/2 or âˆ’1/2So, tan Î¸ =

=> Î¸ = tan

^{âˆ’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}âˆ’ 64x = 0=> x(x

^{2}âˆ’ 64) = 0=> x = 0 or x

^{2}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 m

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

m

_{1}= 3x^{2}/4y = âˆžm

_{2}= 16/y = âˆžSo, tan Î¸ = âˆž

=> Î¸ = Ï€/2

When x = Â±8, then y = Â±16.

m

_{1}= 3x^{2}/4y = 3 or âˆ’3m

_{2}= 16/y = 1 or âˆ’1So, tan Î¸ =

=> Î¸ = tan

^{âˆ’1}(1/2)

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

^{2}+ y

^{2}â€“ 4x â€“ 1 = 0 and x

^{2}+ y

^{2}â€“ 2y â€“ 9 = 0

**Solution:**

First curve is x

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

=> 2x + 2y (dy/dx) â€“ 4 = 0

=> m

_{1}= dy/dx = (2â€“x)/ySecond curve is x

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

=> 2x + 2y (dy/dx) â€“ 2 (dy/dx) = 0

=> m

_{2}= dy/dx = â€“x/(yâ€“1)First curve can be written as,

=> (x â€“ 2)

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

=> x

^{2}+ y^{2}â€“ 4x â€“ 1 â€“ x^{2}â€“ y^{2}+ 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 = 0â‡’ (x â€“ 2)

^{2}(1 + 4) â€“ 5 = 0â‡’ 5(x â€“ 2)

^{2}â€“ 5 = 0â‡’ (x â€“ 2)

^{2}= 1â‡’ x = 3 or x = 1

So, when x = 3 then y = 6 â€“ 4 = 2

m

_{1}= (2â€“x)/y = (2â€“3)/2 = â€“1/2m

_{2}= â€“x/(yâ€“1) = â€“3/(2â€“1) = â€“3So, tan Î¸ == 1

=> Î¸ = Ï€/4

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

m

_{1}= (2â€“x)/y = (2â€“1)/(â€“2) = â€“1/2m

_{2}= â€“x/(yâ€“1) = â€“1/(â€“2â€“1) = 1/3So, tan Î¸ == 1

=> Î¸ = Ï€/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 = â€“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 = â€“2x/2y = â€“x/ySolving (1) and (2), we get,

=> x

^{2}/a^{2}+ (ab â€“ x^{2})/b^{2}= 1=> x

^{2}b^{2}â€“ a^{2}x^{2}= a^{2}b^{2}â€“ a^{3}b=> x

^{2}==> x =

From (2), we get, y

^{2}==> y =

So, m

_{1}= â€“b^{2}x/a^{2}y ==

m

_{2}= â€“x/y ==

Therefore, tan Î¸ =

=> tan Î¸ =

=> tan Î¸ =

=> Î¸ = tan

^{â€“1}((aâ€“b)/âˆšab)

**(vi) x**^{2} + 4y^{2} = 8 and x^{2} â€“ 2y^{2} = 2

^{2}+ 4y

^{2}= 8 and x

^{2}â€“ 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 = â€“2x/8y = â€“x/4ySecond curve is x

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

=> 2x â€“ 4y (dy/dx) = 0

=> m

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

6y

^{2}= 6 => y2 = Â±1x

^{2}= 2 + 2 => x = Â±2So, tan Î¸ =

=> Î¸ = tan

^{â€“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}âˆ’ 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}= âˆžtan Î¸ == âˆž

=> Î¸ = Ï€/2

So, when x = 18 and y = 12

m

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

=> Î¸ = tan

^{âˆ’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 = (1â€“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}â€“ x = 0=> x = 0 or x = 1

And y = 0 or y = Â±1.

When x = 0, y = 0, m

_{1}= âˆž and m_{2}= âˆžtan Î¸ =

=> Î¸ = Ï€/2

When x = 1 and y = Â±1, m

_{1}= 0 and m_{2}= 1/2tan Î¸ =

=> Î¸ = tan

^{âˆ’1}(1/2)

**(ix) y = 4 âˆ’ x**^{2} and y = x^{2}

^{2}and y = x

^{2}

**Solution:**

First curve is y = 4 âˆ’ x

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

=> dy/dx = âˆ’2x

=> m

_{1}= dy/dx = âˆ’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 = Â±âˆš2

And y = 2

So, m

_{1}= âˆ’2x = âˆ’2âˆš2 and m_{2}= 2x = 2âˆš2tan Î¸ =

=> Î¸ = tan

^{âˆ’1}(4âˆš2/7)

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

**(i) y = x**^{3} and 6y = 7 â€“ x^{2}

^{3}and 6y = 7 â€“ 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 â€“ x

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

=> 6y (dy/dx) = â€“ 2x

=> m

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

=> 6y = 7 â€“ x

^{2}=> 6x

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

^{3}= 1So, m

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

_{1}m_{2}= â€“1=> 3 Ã— (â€“1/3) = â€“1

Hence proved.

**(ii) x**^{3} â€“ 3xy^{2} = â€“ 2 and 3x^{2} y â€“ y^{3} = 2

^{3}â€“ 3xy

^{2}= â€“ 2 and 3x

^{2}y â€“ y

^{3}= 2

**Solution:**

First curve is x

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

=> 3x

^{2}â€“ 3y^{2}â€“ 6xy (dy/dx) = 0=> m

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

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

=> 6xy + 3x

^{2}(dy/dx) â€“ 3y^{2}(dy/dx) = 0=> m

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

_{1}m_{2}= â€“1=>= â€“1

Hence proved.

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

^{2}+ 4y

^{2}= 8 and x

^{2}â€“ 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 = â€“x/4ySecond curve is x

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

=> 2x â€“ 4y (dy/dx) = 0

=> m

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

=> x = 4/âˆš3 and y = âˆš2/âˆš3

So, m

_{1}= â€“x/4y = â€“1/âˆš2m

_{2}= x/2y = âˆš2Two curves intersect orthogonally if m

_{1}m_{2}= â€“1=> (â€“1/âˆš2) Ã— âˆš2 = â€“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 = âˆ’2x/4 =âˆ’x/2For x = 2 and y = 1, we have m

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

_{1}m_{2}= â€“1=> 1 Ã— (â€“1) = â€“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 = âˆ’3x^{2}/6 =âˆ’x^{2}/2For x = 1 and y = 1, we have m

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

_{1}m_{2}= â€“1=> 2 Ã— (â€“1/2) = â€“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, 2âˆš2).**

^{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 = âˆ’4x/2y =âˆ’2x/yFor x = 1 and y = 2âˆš2, we have m

_{1}= 4/2âˆš2 = âˆš2 and m_{2}= âˆ’2/2âˆš2 = âˆ’1/âˆš2Two curves intersect orthogonally if m

_{1}m_{2}= 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 = 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 = âˆ’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}= â€“1=> (2/y) Ã— (âˆ’y/x) = â€“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 = âˆ’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}= â€“1=> (1/y) Ã— (âˆ’y/x) = â€“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}âˆ’ 8y^{2}+ 16 = 0=> (y

^{2}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

=> m

_{1}= dy/dx = âˆ’y/xDifferentiating 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,

m

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

_{1}= m_{2}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 y**^{2} = 4x and x^{2} + y^{2} âˆ’ 6x + 1 = 0 touch each other at the point (1, 2).

^{2}= 4x and x

^{2}+ y

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

=> 2x + 2y (dy/dx) âˆ’ 6 = 0

=> m

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

m

_{1}= 2/2 = 1m

_{2}= (3âˆ’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} âˆ’ y^{2}/b^{2} = 1 and xy = c^{2}

^{2}/a

^{2}âˆ’ y

^{2}/b

^{2}= 1 and xy = c

^{2}

**Solution:**

We have,

x

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

=> 2x/a

^{2}âˆ’ (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 = âˆ’y/xFor curves to intersect orthogonally, m

_{1}m_{2}= âˆ’1.=> (b

^{2}x/a^{2}y) (âˆ’y/x) = âˆ’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} âˆ’ y^{2}/B^{2} = 1

^{2}/a

^{2}+ y

^{2}/b

^{2}= 1 and x

^{2}/A

^{2}âˆ’ 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 = âˆ’b^{2}x/a^{2}yAlso, x

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

^{2}âˆ’ (2y/B^{2}) (dy/dx) = 0=> m

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

_{1}m_{2}= âˆ’1.=> (âˆ’b

^{2}x/a^{2}y) (B^{2}x/A^{2}y) = âˆ’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}âˆ’ A^{2}=> a

^{2}âˆ’ b^{2}= A^{2}+ B^{2}

Therefore, a^{2}âˆ’ 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}+ Î»_{1}) + 2y/(b^{2}+ Î»_{1}) (dy/dx) = 0=> m

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

. . . . (2)

Differentiating both sides with respect to x, we get,

=> 2x/(a

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

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

_{1}m_{2}= âˆ’1.=>

=>. . . . (3)

Subtracting (2) from (1) gives,

=>

=>

Putting this value in (3), we get,

=>

=> m

_{1}m_{2}= âˆ’1

Hence proved.

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

^{2}/a

^{2}âˆ’ y

^{2}/b

^{2}= 1, then prove that a

^{2}cos

^{2}Î± âˆ’ b

^{2}sin

^{2}Î± = p

^{2}.

**Solution:**

Suppose (x

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

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

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

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

=>

=> x

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

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

Using (1), we get,

=>

=> a

^{2}cos^{2}Î± âˆ’ b^{2}sin^{2}Î± = p^{2}

Hence proved.

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