Question

Question #4f507

Answer 1

Tangent `x=1`
Normal `y=4`

Explanation:

Given -

`x^2+y^2-3x-8x+18=0`

It is a circle because the coefficient of `x^2` is equal to the coefficient of `y^2`.

If the general form of the equation is -

`ax^2+cy^2+dx+ey+f=0`

Then -

If `a=c` The equation represents a circle.

Examine whether the point `(1, 4)` is on the line.

At `x=1 ; y = 4`

`1^2+4^2-3(1)-8(4)+18=0`

`1 + 16-3-32+18=0`

`35-35=0`

Yes, the given point is on the curve.

Find the slope.

At any given point the slope of the curve is given by its first derivative.

Differentiate the function implicitly.

`2x+2ydy/dx-3-8dy/dx=0`

Solve it for `dy/dx`

`2y.dy/dx-8.dy/dx=-2x+3`

`(2y-8)dy/dx=(-2x+3)`

`dy/dx=((-2x+3))/((2y-8))`

Slope of the curve at `(1, 4)`

At `x=1 ; y = 4`

`dy/dx=(-2(1)+3)/((2(4)-8))=(-2+3)/0=oo`

Hence, tangent is vertical to the x-axis and normal is vertical to the y-axis.

Tangent `x=1`
Normal `y=4`

Look at the graph also