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help maths question

yo can anyone help me with thishard.JPG
Original post by lfs_04_
yo can anyone help me with thishard.JPG

Did you find the equation of the tangent?
Did you find the coordinates of P?
Post your working.
Reply 2
Original post by BuryMathsTutor
Did you find the equation of the tangent?
Did you find the coordinates of P?
Post your working.

So, I drew a line from 0,0 to 2,6 and worked out the gradient of that line, and the tangent is perpendicular to it so the equation of the tangent is y=-2/3x+20/3. From that equation, i then solved for x, which gave me x= 10. Therefore the base of the triangle was 10 and the height was 6, so I then did (10x6)/2 which gave me 30.
Original post by lfs_04_
So, I drew a line from 0,0 to 2,6 and worked out the gradient of that line, and the tangent is perpendicular to it so the equation of the tangent is y=-2/3x+20/3. From that equation, i then solved for x, which gave me x= 10. Therefore the base of the triangle was 10 and the height was 6, so I then did (10x6)/2 which gave me 30.

What did you have as the gradients of OA and AP?
The gradient of your tangent is incorrect.
Reply 4
OA gradient was 3
AP gradient was -1/3
Sorry, i realized my silly mistake- the equation was -1/3x+c, so I substituted 2 into x and then got -2/3, however I forgot i had substituted it and used y= -2/3x as the gradient
Thanks
Original post by lfs_04_
OA gradient was 3
AP gradient was -1/3
Sorry, i realized my silly mistake- the equation was -1/3x+c, so I substituted 2 into x and then got -2/3, however I forgot i had substituted it and used y= -2/3x as the gradient
Thanks


By the way, as a check, you can write down the equation of the tangent at (a,b) without doing any working.

It is ax+by=a2+b2ax+by=a^2+b^2 so in this case it's 2x+6y=402x+6y=40 then for the coordinates of P sub in y=0 and solve 2x=40.


If you use this as a check make sure that you still show a valid method for finding the equation of the tangent.

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