tangent (1 Viewer)

[ozidolroks]

need help with with question e ) i can do the rest but the questions are a follow up of each other

a) Show that the point R(-1.5,-1) is external to the parabola x^2=4y.
b) Find the equation of the chord of contact of the tangents from R.
c) If this chord meets the parabola at P and Q, find the coordinates of P and Q.
d) Find the coordinates where PQ cuts
i) the x axis
ii) the y axis
iii) the directrix ( at the point T)
e) If S is the focus of the parabola show that
ANGLE TSR = ANGLE PRQ = 90 DEGREES

 

[gurmies]

Not sure what is meant by part (a)

(b) xx1 = 2a(y + y1)

-1.5x = 2(y - 1)

-1.5x = 2y - 2

2y = 2 - 1.5x

y = -3x/4 + 1 [1]

(c) If equation of parabola is x^2 = 4y, y = (x^2)/4

(x^2)/4 = -3x/4 + 1

x^2 = -3x + 4

x^2 + 3x - 4 = 0

(x - 1)(x + 4) = 0

x = 1 or/ x = -4

y = 1/4 or/ y = 4

So let P have the co-ordinates (1, 1/4) and Q have the co-ordinates (-4, 4)

(d) Firstly, mPQ = (4 - 1/4)/(-4 -1) = -3/4

y + 4 = -3/4(x + 4)

4y + 16 = -3x - 12

3x + 4y + 28 = 0

(i) When y = 0, 3x = -28

x = -28/3 (-28/3, 0)

(ii) When x = 0, 4y = - 28

y = -7 (0, -7)

(ii) Directrix is y = -1

3x - 4 + 28 = 0

3x = -24

x = - 8 - T(-8, -1)

(e) Focal point is (0, 1)

mTS x mSR = -1

mTS = (-1 -1)/(-8) = 1/4

mSR where R is (-1.5, -1) = (-1 -1)/(-1.5) =/ -4...so i've made a mistake as mSR x mTS =/ -1

P (1, 1/4) and Q(-4, 4) and R (-1.5, -1)

mPR = (-1-1/4)/(-1.5 -1) = 1/2

mQR = (-1 - 4)/(-1.5 + 4) = -2

so mPR x mQR = -1 and thus PQR is a right angle

Not sure where I went wrong above, please somebody correct me

But that's the gist of how you approach that kind of question...

 

[independantz]

e) If S is the focus of the parabola show that
ANGLE TSR = ANGLE PRQ = 90 DEGREES

P(-4,4), Q(1,1/4), R(-3/2,-1)

RTP:M_PRxM_RQ=-1

M_PR=[4+1]/[-4+(3/2)]=5/[-5/2]=-2
M_QR=[(1/4)+1]/[1+3/2]=1/2

LHS=M_PRxM_RQ=2x(-1/2)=-1= RHS

The other one is done similarly.