Question

Find the slope of the chord of parabola y^2=4x whose midpoint is (1,1). Help with explanation

Answer 1

Slope = 2 for chord of parabola y^2=4x whose midpoint is (1,1).

Step-by-step explanation:

Let say coordinates of chord are

(x₁ , y₁ )   & (x₂ , y₂)

Midpoint = ( 1. 1)

= (x₁ + x₂)/2 = 1

=> x₁ + x₂ = 2

y₁ + y₂ = 2

y₁² = 4x₁

y₂² = 4x₂

Adding both

y₁²  + y₂²  = 4(x₁ + x₂)

=> y₁² + (2 - y₁)² = 4 * 2

=> 2y₁² - 4y₁ + 4 = 8

=> y₁² - 2y₁ - 2=  0

y₁  = 1 + √3     or 1 - √3

y₂ = 1  -  √3    or  1 +√3

y₁² = 4x₁

=>( 1 + √3)² = 4x₁

=> x₁ = 1  + √3/2

similarly  x₂ = 1  - √3/2

1  + √3/2 ,  1 + √3  

&  1  - √3/2 , 1 - √3  

Slope = ( 2√3 / √3)  = 2

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The slope of the chord of parabola y^2=4x whose midpoint is (1,1)

CALCULATION

$\sf{Let \: (x_1,y_1) \: and \: (x_2,y_2) \: be \: the \: two \: points}$

Then

$\sf{{y_1}^{2} = 4 x_1 \ \: .......(1) \: }$

$\sf{ {y_2}^{2} = 4 x_2 \: \: \: \: ........(2)\: }$

On substraction

$\sf{{y_2}^{2} - {y_1}^{2} =4 x_2 - 4 x_1 }$

$\implies \: \sf{({y_2} + {y_1} )( {y_2} - {y_1} )\: \: =4( x_2 - x_1) }$

$\implies \: \displaystyle \: \sf{ \frac{( {y_2} - {y_1} )}{( x_2 - x_1)} \: \: \sf{\: \: =\frac{4}{({y_2} + {y_1} )} }.......(3)}$

Again

$\sf{(1,1) \: is \: the \: midpoint \: of \: the \: line \: joining\: (x_1,y_1) \: and \: (x_2,y_2) \:}$

So

$\implies \: \displaystyle \: \sf{ \frac{( x_2 + x_1)}{2} = 1 \: \: \: and \: \: \frac{( {y_2} + {y_1} ) }{2} = 1\: }$

$\implies \: \displaystyle \: \sf{ ( x_2 + x_1) = 2 \: \: \: and \: \: ( {y_2} + {y_1} ) = 2\: } \: ......(4)$

$\sf{Therefore \: the \: slope \: of \: the \: chord \: joining \: the}$

$\sf{\: (x_1,y_1) \: and \: (x_2,y_2) } \: is$

$= \displaystyle \: \sf{ \frac{( {y_2} - {y_1} )}{( x_2 - x_1)} \: \: }$

$= \displaystyle \sf{ \frac{4}{({y_2} + {y_1} )} } \: \: \: \: \: \: \: \: \: using \: \: \: (3)$

$= \displaystyle \: \sf{ \frac{4}{2} \: }$

$= \displaystyle \: \sf{2}$

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