Question

The line 2x - y + 4 = 0 cuts the parabola y2 = 8x in P and Q. The midpoint of PQ is

Answer 1

EXPLANATION.

The line 2x - y + 4 = 0 cuts the parabola y² = 8x.

As we know that,

We can write equation as,

⇒ 2x - y + 4 = 0.

⇒ y = 2x + 4.

Put the value of y = 2x + 4 in equation y² = 8x, we get.

⇒ y² = 8x.

⇒ (2x + 4)² = 8x.

As we know that,

Formula of :

⇒ (a + b)² = a² + b² + 2ab.

Using this formula in the equation, we get.

⇒ (2x)² + (4)² + 2(2x)(4) = 8x.

⇒ 4x² +16 + 16x = 8x.

⇒ 4x² + 16x - 8x + 16 = 0.

⇒ 4x² + 8x + 16 = 0.

⇒ 4(x² + 2x + 4) = 0.

⇒ x² + 2x + 4 = 0.

As we know that,

Let α₁ and α₂ be the roots of the equation.

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ α₁ + α₂ = -2.

⇒ y² = 8x.

⇒ x = y²/8.

Put the value of x = y²/8 in equation 2x - y + 4 = 0, we get.

⇒ 2x - y + 4 = 0.

⇒ 2(y²/8) - y + 4 = 0.

⇒ y²/4 - y + 4 = 0.

⇒ y² - 4y + 16 = 0.

Let β₁ and β₂ be the roots of the equation.

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ β₁ + β₂ = -(-4)/1 = 4.

⇒ β₁ + β₂ = 4.

Now, we know that,

Formula of :

Midpoint = [(x₁ + x₂)/2, (y₁ + y₂)/2].

⇒ [(α₁ + α₂)/2, (β₁ + β₂)/2].

⇒ [(-2)/(2) , (4)/(2)].

⇒ ( - 1, 2).

The midpoint of PQ = (-1,2).

Answer 2

Solution

Given that the line 2x - y + 4 = 0 cuts the parabola y2 = 8x at P, Q. We need to find the midpoint of PQ.

Substituting y = 2x + 4 in the equation of parabola.

⇒ (2x + 4)2 = 8x

⇒ 4x^2 + 16x + 16 = 8x

⇒ 4x^2 + 8x + 16 = 0

⇒ x^2 + 2x + 4 = 0

Let x1, x2 be the roots. Then, x1 + x2 = - 2

Now substituting

$x = \frac{y {}^{2} }{8}$

in the equation of the line we get,

${ \implies}2 (\frac{y {}^{2} }{8} ) - y + 4 \\ { \implies}\frac{y {}^{2} }{4} - y + 4$

⇒ y2 - 4y + 16 = 0

Let y1, y2 be the roots. Then, y1 + y2 = 4

Let us assume P be (x1, y1) and Q be (x2, y2) and R be the midpoint of PQ.

⇒

$r = \frac{x \frac{}{1} + x \frac{}{2} }{2} \\ r = ( \frac{ - 2}{2} , \frac{4}{2} )$

⇒ r = (- 1, 2)