Question

From an external point P, a pair of tangents is drawn to the parabola $\sf{y^2=4x}$. If $\sf{\theta_1\:and\:\theta_2}$ are the inclinations of these tangents with the x-axis such that $\sf{\theta_1+\theta_2=\dfrac{\pi}{4}}$, then find the locus of P. ​

Answer 1

Answer:

y = x - 1

Step-by-step explanation:

Let the locus of P be ( h , k ) .

Given :

Equation of parabola :

y² = 4 x

Comparing it with standard equation of parabola i.e.

y² = 4 a x

On comparing we get :

a = 1

We know :

Equation of tangent to parabola whose slope is m

y = m x + a / m  [ a = 1 ]

y = m x + 1 / m

Since ( h , k ) lie on point so it will satisfy the equation :

k = h m + 1 / m

= > m² h -  m k + 1 = 0

Here we will get two value of m let say they are m₁ and m₂ :

Their sum m₁ + m₂ = - b / a = k / h  ( i )

Their product m₁ m₂ = c / a = 1 / h ... ( ii )

We have also given :

Ф₁ + Ф₂ = π / 4

Taking tan both side :

tan ( Ф₁ + Ф₂ ) = tan ( π / 4 )

Using formula :

tan ( A + B ) = tan A + tan B / 1 - tan A tan B

Replacing tan A by m₁  and tan B by m₂ :

( m₁ + m₂  ) / ( 1 - m₁ m₂ ) = tan π / 4

Using  ( i ) and ( ii ) we get :

( k / h ) / ( 1 - 1 / h ) = 1    [ tan π / 4 = 1 ]

( k / h ) / ( ( h - 1 ) / h ) = 1

k / ( h - 1 ) = 1

k = h - 1

Replacing in form of x and y

y = x - 1

Therefore , the locus of P is y = x - 1 .

Answer 2

Answer:

y=x-1

Step-by-step explanation:

hope it is helpful to you..