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# Need a more effective method watch

1. The initial question was to find the equation of the normal to the curve when A is 45°which gives 6y= 4x + 5√2

x = 2Cos(A)
y = 3Sin(A)
Cos^2(A) + Sin^2(A) = 1
x^2 /4 + y^2/9 = 1
2nd part asked another set of points where the normal and curve meet.
So i made y subject of formula for both cartesian equation and solved for x and ended up with a very big quadratic eqn which forced me to use the formula. Is there a faster and less time consuming method? ( it took me about 7-8mins to complete it which i guess is a lot for A level exams)
2. (Original post by Bilbao)
The initial question was to find the equation of the normal to the curve when A is 45°which gives 6y= 4x + 5√2

x = 2Cos(A)
y = 3Sin(A)
Cos^2(A) + Sin^2(A) = 1
x^2 /4 + y^2/9 = 1

So i made y subject of formula for both cartesian equation and solved for x and ended up with a very big quadratic eqn which forced me to use the formula. Is there a faster and less time consuming method? ( it took me about 7-8mins to complete it which i guess is a lot for A level exams)
dy/dx=dy/dA*dA/dx= -3/2*cot(A)

Evaluate x, y, and dy/dx at A=45 then just plug em into y-b=m(x-a) [where m is the normal gradient in this case]
3. (Original post by RDKGames)
dy/dx=dy/dA*dA/dx= -3/2*cot(A)

Evaluate x, y, and dy/dx at A=45 then just plug em into y-b=m(x-a) [where m is the normal gradient in this case]
Ah sorry i missed the actual question, i actually did what you wrote above to find the equation of the nornal to the curve. The 2nd part of the question asked to find another point where the normal and curve meet again. And i used the method which i wrote above. Im looking for a more efficient way to find another point other than equating the two cartesian equations above and then solving for x.
4. (Original post by Bilbao)
Ah sorry i missed the actual question, i actually did what you wrote above to find the equation of the nornal to the curve. The 2nd part of the question asked to find another point where the normal and curve meet again. And i used the method which i wrote above. Im looking for a more efficient way to find another point other than equating the two cartesian equations above and then solving for x.
You can sub in x=f(A) and y=g(A) into the line’s equation. Solve for A not equal to 45.
Not sure whether faster as I cant test it atm, but seems like a nice alternative
5. (Original post by RDKGames)
You can sub in x=f(A) and y=g(A) into the line’s equation. Solve for A not equal to 45.
Not sure whether faster as I cant test it atm, but seems like a nice alternative
agreed. this is the way i'd do it

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