Tangents and locus qn. (1 Viewer)

[Getteral09]

Hey ppl, Can anyone help me to solve this maths problem?Find the equation of the tangent to the curve x^2=-2y at the point (4,-8). This tangent meets the directrix at point M. Find the coordinates of M.
Thanks,Getteral

 

[pLuvia]

x²=-2y
y=-x²/2
dy/dx=-x
At (4,-8)
dy/dx=-4
Tangent: y+8=-4(x-4)
y=-4x+8

Directrix y=1/2
At M, y=1/2
x=15/8
M(15/8,1/2)

 

[rocky1989]

y=-(1/2)x^2
y'=-x
sub in x=4 to get the gradient of the tangent at (4,-8)
therefore m=-4

then to get the eq of the tangent, use point gradient formula
y+8=-4(x-4)
y=-4x+8

we also know that the directrix has the equation y=-a where we can find a=-(1/2)
therefore directrix: y=(1/2)

now just solve the two simultaneously to get the point M(1.875,0.5)

 

[rocky1989]

Directrix y=-1/2

the porabola is concave down, so the directrix will be above the x-axis

 

[Getteral09]

I don't understand on how you get the a to be -1/2.

thanks

 

[rocky1989]

I don't understand on how you get the a to be -1/2.

thanks

well the porabola is always in the form x^2=4ay
and for this question: 4a=-2
therefore a=-(1/2)