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    HEY i was doing these questions my teacher gave and im confused as to how to do part 2 as i tried it and i keep gettign the wrong answer as im not sure how to go about it , could someone explain it to me ?

    1i) find the equation of the normal of the curve x=4t, y=4/t at the point (8,2)

    and I got y=4x-30 like the mark scheme he have
    by working out that t= 2 and gradient for the normal is 4 and the gradient for the curve was -1/4

    BUT IM UNSURE ABOUT HOW TO DO PART 2 ;(
    (ii)
    find the coordinates of the point where this normal crosses the curve again ?


    thank you would really appreciate the help ;p
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    (Original post by Sadilla)
    HEY i was doing these questions my teacher gave and im confused as to how to do part 2 as i tried it and i keep gettign the wrong answer as im not sure how to go about it , could someone explain it to me ?

    1i) find the equation of the normal of the curve x=4t, y=4/t at the point (8,2)

    and I got y=4x-30 like the mark scheme he have
    by working out that t= 2 and gradient for the normal is 4 and the gradient for the curve was -1/4

    BUT IM UNSURE ABOUT HOW TO DO PART 2 ;(
    (ii)
    find the coordinates of the point where this normal crosses the curve again ?


    thank you would really appreciate the help ;p
    OK so the equation of the normal is y=4x-30 and you're looking for the other point where this intersects the parametric curve (x,y) = (4t, \frac{4}{t}) again.

    One way to do this is to derive the Cartesian equation of the curve then find the points of intersection between this and your line.

    An alternative way would be to substitute x=4t and y=\frac{4}{t} into y=4x-30 then determine the root that isn't t=2. Sub this t back into (x,y)=(4t, \frac{4}{t}) to determine the point.
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    Use the equation y-y1= m(x-x1) to form an equation for the normal ( you have the gradient and y1 and x1)

    Set that equation equal to 16/x as y= 4/ (x/4) = 16/x and solve
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    most people would eliminate t to find the equation of the curve with just y and x, then find the intersection in the usual way.
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    (Original post by Sadilla)
    HEY i was doing these questions my teacher gave and im confused as to how to do part 2 as i tried it and i keep gettign the wrong answer as im not sure how to go about it , could someone explain it to me ?

    1i) find the equation of the normal of the curve x=4t, y=4/t at the point (8,2)

    and I got y=4x-30 like the mark scheme he have
    by working out that t= 2 and gradient for the normal is 4 and the gradient for the curve was -1/4

    BUT IM UNSURE ABOUT HOW TO DO PART 2 ;(
    (ii)
    find the coordinates of the point where this normal crosses the curve again ?


    thank you would really appreciate the help ;p
    You could think about finding the equation of the curve in the form y = f(x) and then solving that and the equation of the normal?
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    thank you, i got the answer if you dont mind could you explain to me why subbing the x values into the normal equation gets you two t values for both the intersection points ?
    and if i did it the other way of working out the equation of the curve would i use t=2 or the other t value which is -1/8 ?
    sorry i feel like im asking dumb question but im just trying to understand it properly


    (Original post by RDKGames)
    OK so the equation of the normal is y=4x-30 and you're looking for the other point where this intersects the parametric curve (x,y) = (4t, \frac{4}{t}) again.

    One way to do this is to derive the Cartesian equation of the curve then find the points of intersection between this and your line.

    An alternative way would be to substitute x=4t and y=\frac{4}{t} into y=4x-30 then determine the root that isn't t=2. Sub this t back into (x,y)=(4t, \frac{4}{t}) to determine the point.
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    (Original post by Sadilla)
    thank you, i got the answer if you dont mind could you explain to me why subbing the x values into the normal equation gets you two t values for both the intersection points ?
    and if i did it the other way of working out the equation of the curve would i use t=2 or the other t value which is -1/8 ?
    sorry i feel like im asking dumb question but im just trying to understand it properly
    By substituting x=4t and y=\frac{4}{t} into the normal, you are essentially substituting the curve into the equation of the line, and hence the leftover equation satisfies the points of intersection for values of t.
    It's sort of like substituting y=2x into y=2x^2 to obtain 2x=2x^2 which is the equation for the two ponts of intersection between y=2x and y=2x^2.
    So then, the equation in t with which you end up with satisfies all the intersection points. You know one of them is for t=2 but this isn't the one we are interested in, so pick the other.

    If you worked out the eq. of the curve in Cartesian, then you wouldn't need to mess around with the parameter t at all since it wouldn't be there. You would then just have your usual x,y to work with. The equation for roots would be in terms of x and you would disregard the solution x=8. This is also the most common approach at A-Level.
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    thank youu soo much it all makes sense now


    (Original post by RDKGames)
    By substituting x=4t and y=\frac{4}{t} into the normal, you are essentially substituting the curve into the equation of the line, and hence the leftover equation satisfies the points of intersection for values of t.
    It's sort of like substituting y=2x into y=2x^2 to obtain 2x=2x^2 which is the equation for the two ponts of intersection between y=2x and y=2x^2.
    So then, the equation in t with which you end up with satisfies all the intersection points. You know one of them is for t=2 but this isn't the one we are interested in, so pick the other.

    If you worked out the eq. of the curve in Cartesian, then you wouldn't need to mess around with the parameter t at all since it wouldn't be there. You would then just have your usual x,y to work with. The equation for roots would be in terms of x and you would disregard the solution x=8. This is also the most common approach at A-Level.
 
 
 
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