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Worksheet: Tangents and Normals to Graphs of Implicit Functions

Q1:

Determine the points on the curve π‘₯ + 𝑦 + π‘₯ + 8 𝑦 = 0 2 2 at which the tangent is perpendicular to the line 7 𝑦 + 4 π‘₯ + 𝑐 = 0 .

  • A ( βˆ’ 3 , βˆ’ 6 ) , ( 4 , βˆ’ 2 )
  • B ( βˆ’ 3 , 6 ) , ( 4 , 2 )
  • C ( 3 , 6 ) , ( βˆ’ 4 , 2 )
  • D ( 3 , βˆ’ 6 ) , ( βˆ’ 4 , βˆ’ 2 )
  • E ( βˆ’ 3 , βˆ’ 6 ) , ( βˆ’ 4 , 2 )

Q2:

Find the equation of the tangent to the curve 9 π‘₯ βˆ’ 6 π‘₯ + 6 π‘₯ βˆ’ 𝑦 βˆ’ 𝑦 + 2 = 0 3 2 2 at the point ( 0 , 1 ) .

  • A 𝑦 βˆ’ 3 π‘₯ βˆ’ 1 = 0
  • B 𝑦 + π‘₯ 2 βˆ’ 1 = 0
  • C 𝑦 + π‘₯ 3 βˆ’ 1 = 0
  • D 𝑦 βˆ’ 2 π‘₯ βˆ’ 1 = 0

Q3:

Find the equation of the tangent to that has gradient .

  • A
  • B
  • C
  • D

Q4:

Do the curves 9 𝑦 βˆ’ 8 𝑦 = 6 π‘₯ 4 and βˆ’ 5 π‘₯ βˆ’ 3 𝑦 = βˆ’ 4 π‘₯ 2 intersect orthogonally at the origin?

  • Ayes
  • Bno

Q5:

At a point on the curve π‘₯ + 3 π‘₯ + 𝑦 + 5 𝑦 + 4 = 0 2 2 with π‘₯ < 0 , 𝑦 < 0 , the tangent makes an angle of 9 πœ‹ 4 with the positive π‘₯ -axis. Find the equation of the tangent at that point.

  • A βˆ’ π‘₯ + 𝑦 βˆ’ 4 = 0
  • B π‘₯ + 𝑦 + 4 = 0
  • C βˆ’ π‘₯ + 𝑦 + 2 = 0
  • D βˆ’ π‘₯ + 𝑦 βˆ’ 2 = 0

Q6:

Find the points on the curve 5 π‘₯ βˆ’ 8 π‘₯ 𝑦 + 4 𝑦 = 4 2 2 at which the tangent is parallel to the 𝑦 axis.

  • A ( 1 , 1 ) , ( βˆ’ 1 , βˆ’ 1 )
  • B ( 2 , 2 )
  • C ( 1 , 1 )
  • D ( 2 , 2 ) , ( βˆ’ 2 , βˆ’ 2 )

Q7:

Find the points that lie on the curve 2 π‘₯ βˆ’ π‘₯ 𝑦 + 2 𝑦 βˆ’ 4 8 = 0 2 2 at which the tangent is parallel to line 𝑦 = βˆ’ π‘₯ .

  • A ( 4 , βˆ’ 4 ) , ( βˆ’ 4 , 4 )
  • B ( 4 . 7 6 , 2 . 8 5 ) , ( βˆ’ 4 . 7 6 , βˆ’ 2 . 8 6 )
  • C ( 4 . 7 6 , 2 . 8 5 ) , ( βˆ’ 4 . 7 6 , βˆ’ 4 . 7 6 )
  • D ( 4 , 4 ) , ( βˆ’ 4 , βˆ’ 4 )

Q8:

A tangent to π‘₯ + 𝑦 = 7 2 2 2 forms an isosceles triangle when taken with the positive π‘₯ - and 𝑦 -axes. What is the equation of this tangent?

  • A βˆ’ π‘₯ + 𝑦 βˆ’ 1 2 = 0
  • B βˆ’ π‘₯ + 𝑦 = 0
  • C π‘₯ + 𝑦 = 0
  • D π‘₯ + 𝑦 βˆ’ 1 2 = 0

Q9:

The point ( βˆ’ 5 , βˆ’ 2 ) lies on the curve π‘₯ + 𝑦 βˆ’ 3 π‘˜ π‘₯ + 7 = 0 2 2 . Find π‘˜ and the equation of the tangent to the curve at this point.

  • A π‘˜ = βˆ’ 1 2 5 , equation of the tangent: 7 π‘₯ 1 0 + 𝑦 + 3 2 = 0
  • B π‘˜ = 1 2 5 , equation of the tangent: 4 3 π‘₯ 1 0 + 𝑦 + 4 7 2 = 0
  • C π‘˜ = βˆ’ 1 2 5 , equation of the tangent: βˆ’ 7 π‘₯ 1 0 + 𝑦 + 1 1 2 = 0
  • D π‘˜ = βˆ’ 1 2 5 , equation of the tangent: 7 π‘₯ 1 0 + 𝑦 + 1 1 2 = 0

Q10:

The tangent at ( βˆ’ 2 , 2 ) to the curve π‘₯ + π‘₯ 𝑦 + 5 π‘₯ + 5 𝑦 = 0 3 2 makes a positive angle with the positive π‘₯ -axis. Find this angle.

  • A 6 0 ∘
  • B 3 0 ∘
  • C 9 0 ∘
  • D 1 3 5 ∘
  • E 4 5 ∘

Q11:

Determine the points on a curve π‘₯ + 𝑦 = 4 5 2 2 at which the tangent to the curve is perpendicular to the straight line 𝑦 = 2 π‘₯ + 1 2 .

  • A ( 6 , 3 ) , ( βˆ’ 6 , βˆ’ 3 )
  • B ( 3 , βˆ’ 6 ) , ( βˆ’ 3 , 6 )
  • C ( 6 , βˆ’ 3 ) , ( βˆ’ 6 , 3 )
  • D ( 3 , 6 ) , ( βˆ’ 3 , βˆ’ 6 )

Q12:

Find the equation of the tangent to the curve at the point .

  • A
  • B
  • C
  • D

Q13:

Find the equation of the tangent to the curve π‘₯ + 𝑦 + 4 π‘₯ βˆ’ 5 𝑦 βˆ’ 5 = 0 2 2 at the point ( 1 , 0 ) .

  • A 7 𝑦 βˆ’ 4 π‘₯ βˆ’ 7 = 0
  • B 7 𝑦 + 4 π‘₯ βˆ’ 7 = 0
  • C 6 𝑦 + 5 π‘₯ + 5 = 0
  • D 5 𝑦 βˆ’ 6 π‘₯ + 6 = 0

Q14:

Find the equations of the tangent lines of the curves and at the intersection points, and state whether the curves intersect orthogonally or not.

  • A , , , , intersect orthogonally
  • B , , , , do not intersect orthogonally
  • C , , , , do not intersect orthogonally
  • D , , , , intersect orthogonally

Q15:

The two curves ( π‘₯ βˆ’ π‘Ž ) + 𝑦 = 5 0   and ( π‘₯ + π‘Ž ) + 𝑦 = 5 0   intersect orthogonally. Find all the possible values of π‘Ž .

  • A √ 5 0 , βˆ’ √ 5 0
  • B25
  • C50
  • D 5 , βˆ’ 5

Q16:

Determine the gradient of the tangent to at its intersection with the -axis.

  • A ,
  • B ,
  • C ,
  • D ,
  • E ,

Q17:

At a point on the curve π‘₯ + 5 π‘₯ + 𝑦 βˆ’ 𝑦 βˆ’ 6 = 0 2 2 with π‘₯ < 0 , 𝑦 < 0 , the tangent makes an angle of 7 πœ‹ 4 with the positive π‘₯ -axis. Find the equation of the normal at that point.

  • A π‘₯ + 𝑦 + 7 = 0
  • B βˆ’ π‘₯ + 𝑦 + 7 = 0
  • C βˆ’ π‘₯ + 𝑦 + 3 = 0
  • D βˆ’ π‘₯ + 𝑦 βˆ’ 3 = 0

Q18:

Find the area of the triangle bounded by the π‘₯ -axis, the tangent, and the normal to the curve π‘₯ + 5 𝑦 = 1 5 2 2 at the point ( 9 , 2 ) to the nearest thousandth.

Q19:

Find the equation of the normal to the curve of the function π‘₯ 6 𝑦 = βˆ’ 6 𝑦 6 π‘₯ s i n c o s at ο€» πœ‹ 4 , πœ‹ 2  .

  • A βˆ’ 1 2 π‘₯ + 𝑦 + 5 πœ‹ 2 = 0
  • B 1 2 π‘₯ + 𝑦 βˆ’ 7 πœ‹ 2 = 0
  • C βˆ’ π‘₯ 1 2 + 𝑦 βˆ’ 2 3 πœ‹ 4 8 = 0
  • D π‘₯ 1 2 + 𝑦 βˆ’ 2 5 πœ‹ 4 8 = 0

Q20:

Find the equation of the normal to the curve π‘₯ 𝑦 βˆ’ 4 π‘₯ + 2 𝑦 βˆ’ 2 0 = 0 2 2 at the point ( 1 , 4 ) .

  • A 𝑦 βˆ’ 2 π‘₯ 5 βˆ’ 1 8 5 = 0
  • B 𝑦 + 1 4 π‘₯ 5 βˆ’ 3 4 5 = 0
  • C 𝑦 + 5 π‘₯ 2 βˆ’ 1 3 2 = 0
  • D 𝑦 βˆ’ 5 π‘₯ 1 4 βˆ’ 5 1 1 4 = 0
  • E 𝑦 + 1 5 π‘₯ 4 βˆ’ 3 1 4 = 0

Q21:

The tangent at ( 1 , 1 ) to the curve 5 π‘₯ 𝑦 + 2 𝑦 π‘₯ = 7 makes a positive angle with the positive π‘₯ -axis. Find this angle.

  • A 6 0 ∘
  • B 3 0 ∘
  • C 9 0 ∘
  • D 4 5 ∘
  • E 1 3 5 ∘

Q22:

Find the equation of the tangent to the curve π‘₯ 6 𝑦 = βˆ’ 8 𝑦 6 π‘₯ s i n c o s at ο€» πœ‹ 4 , πœ‹ 2  .

  • A π‘₯ 1 6 + 𝑦 βˆ’ 3 3 πœ‹ 6 4 = 0
  • B 1 6 π‘₯ + 𝑦 βˆ’ 9 πœ‹ 2 = 0
  • C βˆ’ π‘₯ 1 6 + 𝑦 βˆ’ 3 1 πœ‹ 6 4 = 0
  • D βˆ’ 1 6 π‘₯ + 𝑦 + 7 πœ‹ 2 = 0

Q23:

At the point ( 0 , βˆ’ 2 ) , determine the equation of the normal to the curve represented by the equation 6 π‘₯ + 2 π‘₯ + 2 π‘₯ βˆ’ 9 𝑦 βˆ’ 8 𝑦 + 2 0 = 0 3 2 2 .

  • A 𝑦 + π‘₯ 1 8 + 2 = 0
  • B 𝑦 + π‘₯ 1 4 + 2 = 0
  • C 𝑦 βˆ’ 1 8 π‘₯ + 2 = 0
  • D 𝑦 βˆ’ 1 4 π‘₯ + 2 = 0

Q24:

Find the points of βˆ’ π‘₯ + 2 𝑦 = βˆ’ 4 2 2 where the angle between the tangent and the positive π‘₯ -axis has cosine 4 5 .

  • A ( βˆ’ 6 , 4 ) , ( 6 , βˆ’ 4 )
  • B ο€Ό 3 8 , 1 4  , ο€Ό βˆ’ 3 8 , βˆ’ 1 4 
  • C ο€Ό βˆ’ 3 8 , 1 4  , ο€Ό 3 8 , βˆ’ 1 4 
  • D ( 6 , 4 ) , ( βˆ’ 6 , βˆ’ 4 )

Q25:

Find the equation of the tangent to the curve 𝑦 = 4 π‘₯ + 4 π‘₯ + 3 3 2 at π‘₯ = 2 .

  • A 2 7 π‘₯ 2 0 + 𝑦 βˆ’ 5 7 1 0 = 0
  • B 2 0 π‘₯ 2 7 + 𝑦 βˆ’ 1 2 1 2 7 = 0
  • C βˆ’ 2 7 π‘₯ 2 0 + 𝑦 βˆ’ 3 1 0 = 0
  • D βˆ’ 2 0 π‘₯ 2 7 + 𝑦 βˆ’ 4 1 2 7 = 0