Question

7.
The focus and directrix of a parabola are (1, --1)
and x + y + 3 = 0. Its vertex is
(a) (GH) (0) (1 : 1) (1 : 1) (a) (12)​

Answer 1

❏ QuesTion :-

→ Given above ↑

❏ Answer :-

→ Vertex of this parabola is $(\frac{ 1}{4} , \: \frac{ - 7}{4} ) \:$

✦To Find :-

→ Vertex of parabola .

❏ Step - by - step explanation :-

Given that ,

→ Focus of parabola = ( 1 , -1) and its directrix is

x + y + 3 = 0 .....eq.(1)

◗ Slope of equation of directrix

$\to \sf \: 1 + \frac{dy}{dx} = 0 \\ \\ \to \sf \: slpe \{ \frac{dy}{dx} \} = - 1 \\ \\ \to \: \sf \: m_1 \: \{ \frac{dy}{dx} \}\: = - 1 \\ \\ \because \sf \: \: m_1 \: \times m_2 = - 1 \\ \\ \to \sf \: m_2 = 1$

We know that the products of the slopes of the perpendicular lines is - 1 .

$\sf \: We \: know \: that \: the \: equation \: of \: the \: line \\ \sf passing \: through \: the \: point \: (x_1, y_1) \: and \: \\ \sf having \: slope \: m \: is \: - \\ \sf (y - y_1 ) = m(x - x_1)</p><p> \: \\ \\ \to \sf \: y - ( - 1) = 1(x - 1) \\ \\ \to \sf \: y -x + 2 = 0 \: \: .....eq.(2)$

On solving equations (2) and (1)

Adding these -

$\to \sf 2y + 5 = 0 \\ \\ \to \boxed{ \sf \: y = \frac{ - 5}{2} } \\ \\ \sf \: put \: this \: in \: equation \: (1) \\ \\ \to \sf \: x - \frac{5}{2} + 3 = 0 \\ \\ \to \boxed{ \: x = \frac{ - 1}{2} } \\ \\ \sf \: We \: know \: that \: vertex \: is \: the \: mid \: \\ \sf point \: of \: focus \: and \: point \: of \: intersection \\ \sf \: of \: axis \: and \: directrix. \: \\ \sf Let \: (a, \: b) \: be \: the \: vertex \: of \: the \: parabola. \\ </p><p>⇒ \sf \: (a, \: b) = \bigg( \frac{ \frac{ - 1}{2} +1}{2} , \: \frac{ \frac{ - 5}{2} - 1}{2} \bigg)</p><p> \\ \\ \to \: (a, \: b) \: = (\frac{ 1}{4} , \: \frac{ - 7}{4} ) \:</p><p>$

Check given points of focus again plz

Answer 2

Question :--- The focus and directrix of a parabola are (1, -1)and x + y + 3 = 0. Its vertex is ?

Concept used :---

→ Vertex of a parabola is the midpoint of the focus and point of intersection of axis and directrix.

→ Since axis and parabola are perpendicular. the line joining focus and intersection point is perpendicular to directrix.

→ The slope m of a line is one of the elements in the equation of a line when written in the "slope and intercept" form: y = mx+b.

The m in the equation is the slope of the line described here.

→ The slopes of perpendicular lines are opposite reciprocals of each other. Their product is -1 ....

_____________________________

Solution :----

Given, Directrix = x+y+3 = 0

→ y = -x - 3 = mx+c

→ m(slope) = (-1) = m1

Now, We know m1* m2=-1

→ (-1)*m2 = (-1)

∴ m2 = 1

Now, let the Point of intersection be (x , y)

So

→ [ y-(-1) ] / [ x -1] = 1

→ (y+1) = (x-1)

→ x - y - 2 = 0 ----------- Equation (1)

Now, it is given that, (x,y) lies on directrix ,

So,

→ x + y + 3 = 0 ------------ Equation (2)

Adding Equation (1) and (2) now , we get,

→ (x-y-2) + (x+y+3) = 0

→ 2x +1 = 0

→ 2x = (-1)

→ x = (-1/2)

Putting value of x in Any Equation now, we get,

→ (-1/2) + y + 3 = 0

→ y = -3 + 1/2

→ y = (-5/2)

____________________________

Now , Vertex or Mid-Points are :----

→ [ (-1/2 + 1)/2 ] ,,,,,,, [ (-5/2 - 1 )/2 ]

→ [ 1/2*2 ] ,,,,,,, [ -7/2*2 ]

→ [ 1/4 ] ,,,,,, [ -7/4 ]

Hence , Points of Vertex will be (1/4) and (-7/4) ..