Find the family of curves whose tangents form an angle π 4 with the hyperbola x y = C , C > 0 .

Let m 1 = d y d x (slope of tangent to the curve) x y = C . . . i Differentiating w.r.t., x , we get x d y d x + y = 0 ⇒ d y d x = - y x = m 2 . . . ii ∵ tan θ = m 1 - m 2 1 + m 1 m 2 Given θ = π 4 ∴ tan π 4 = d y d x - - y x 1 - d y d x y x ⇒ ± 1 = d y d x - - y x 1 - d y d x y x . . . . A Taking positive sign, 1 - y x d y d x = d y d x + y x ⇒ d y d x = 1 - y x 1 + y x ⇒ d y d x = x - y x + y ⇒ d y d x = x - C x x + C x . (using i ) ⇒ d y d x = x 2 - C x 2 + C . . . iii ⇒ d y = x 2 + C - 2 C x 2 + C d x ⇒ d y = 1 d x - 2 C x 2 + C d x Integrating both the sides ∫ d y = ∫ d x - 2 C · ∫ d x C 2 + x 2 y = x - 2 C · 1 C · tan - 1 x C + k y = x - 2 C tan - 1 x C + k . Taking negative sign in A - 1 + y x d y d x = d y d x + y x ⇒ y x - 1 d y d x = 1 + y x ⇒ d y d x = y x + 1 y x - 1 ⇒ d y d x = y + x y - x ⇒ d y d x = C x + x C x - x . (using i ) ⇒ d y d x = C + x 2 C - x 2 ⇒ d y = - x 2 - C + 2 C x 2 - C d x ⇒ d y = - d x + 2 C x 2 - C d x Integrating both the sides ∫ d y = - ∫ d x + 2 C · ∫ d x x 2 - C 2 ⇒ y = - x + 2 C 1 2 C log x - C x + C + k 1 ⇒ y = - x + C log x - C x + C + k 1

E M B I B E

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 64

MEDIUM

JEE Main

IMPORTANT

Earn 100

Find the curve for which the intercept cut off by a tangent on the $x - a x i s$ is equal to four times the ordinate of the point of contact.

Important Questions on Differential Equations - II

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 65

Show that equation to the curve such that the

y -

intercept cut off by the tangent at an arbitrary point is proportional to the square of the ordinate of the point of tangency is in the form

\frac{a}{x} + \frac{b}{y} = 1

.

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 66

Find the equation of the curve which is such that the area of the rectangle constructed on the abscissa of any point and the initial ordinate of the tangent at the point is a constant

k^{2}

.

MEDIUM

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 67

Show that the equation of the curve passing through the point

(2, 1)

and for which the sum of the subtangent and the abscissa is equal to

1

is

y (x - 1) = 1

.

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 68

Find the family of curves whose tangents form an angle

\frac{π}{4}

with the hyperbola

x y = C, C > 0

.

MEDIUM

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 69

Prove that the family of curves for which subnormal is a constant in a parabola.

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 70

Find the equation of the curve whose slope at

x = 0

is

3

which passes through the point

(0, 1)

satisfying the differential equation

(x^{2} + 1) \frac{d^{2} y}{d x^{2}} = 2 x \frac{d y}{d x}

.

MEDIUM

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 71

Through any point

(x, y)

of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. Given that it divides the rectangle formed by the two lines and the axes into two areas one of which is twice the other, show that such a curve is a parabola.

MEDIUM

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 10 - Differential Equations - II>EXERCISE>Q 72

If the length of the portion of the normal intercepted between a curve and the